0. What do you know about acid by Arrhenius, Brondted Lowry and Lewis Jawab: Arrhenius : asam adalah zat + yang mengandung ion H jika dilarutkan dalam air Bronsted Lowry : asam adalah zat yang mampu mendonorkan proton Lewis : asam adalah zat yang menerima pasangan elektron 1.
Write the chemical equation for the autoionization of water and the equilibrium law for K w? Jawab: Autoionisation of water is H2O + H + OH + H2O H + OH K = + K [H2O]= [H ] [OH ] + Kw = [H ] [OH ]
2.
3.
How are acidic, basic, and neutral solutions in water defined + a. in terms of [H ] and [OH ] and b. in terms of pH ? Jawab: + a. Acidic = [H ] > [OH ] + Basic = [H ] < [OH ] + Neutral = [H ] = [OH ] b. pH 7 neutral ( 298 K ) pH < 7 acid ( 298K ) pH < 7 basic ( 298 K ) At the temperature of the human body, o -14 37 C, the value of Kw is 2.4 x 10 . + Calculate the [H ], [OH ], pH and pOH of pure water at this temperature. What is the relation between pH, pOH, and Kw at this temperature? Is water neutral at this temperature? Jawab: o d1 : T = 37 C = 310 K -14 Kw = 2.4 x 10 + d2 : [H ], [OH ], pH, pOH d3 :
+
-
H2O ↔ H + OH
Kw = + Kw = [H ] [OH ] -14 2.4 x 10 = x2 x = -7 x = 1.549 x 10 + -7 [H ] = 1.549 x 10 -7 [OH ] = 1.549 x 10 + pH = - log [H ] -7 = - log (1.549 x 10 ) = 7 – 7 – log log 1.549 = 6.8099 pOH = - log [OH ] -7 = - log (1.549 x 10 ) = 7 – 7 – log log 1.549 = 6.8099 Pada suhu ini, pH = pOH yang merupakan ½ pKw. Air bersifat netral 4. Deuterium oxide, D2O, ionizes like water. At 20°C its Kw, or ion product constant analogous to that of water, is -16 + 8.9 x 10 . Calculate [D ] and [OD ] in deuterium oxide at 20°C. Calculate also the pD and the pDO. Jawab: d1 = K w = 8.9 x 10-16 T = 20°C + D2O→D + OD + d2 = [D ], [OD ], pD, pOD = ...? d3
+
-
= [D ] = [OD ]
= = -16 = 2.98 x 10 = 8 - log 2.98 = 7.53 = 8 - log 2.98 = 7.53
pD pOD +
5. Calculate the H concentration in each of the following solutions in which the hydroxide ion concentrations are : a. 0.0024 M -5 b. 1.4 x 10 M -9 c. 5.6 x 10 M
-13
d. 4.2 x 10 M Jawab: -3 a. [OH ] = 2.4 x 10 M pOH = - log [OH ] -3 = - log 2.4 x 10 = 3 - log 2.4 pH = 14- (3- log 2.4 ) = 11 + log 2.4 = 11.38 + -11.38 [H ] = 10 -12 = 4.168 x 10 M -5 b. [OH ] = 1.4 x 10 M pOH = - log [OH ] -5 = - log 1.4 x 10 = 5 - log 1.4 pH = 14- (5- log 1.4 ) = 9 + log 1.4 = 9.146 + -9.146 [H ] = 10 -10 = 7.145 x 10 M -
-9
c. [OH ] = 5.6 x 10 M pOH = - log [OH ] -9 = - log 5.6 x 10 = 9 - log 5.6 pH = 14- (9- log 5.6 ) = 5 + log 5.6 = 5.748 + -5.748 [H ] = 10 -6 = 1.786 x 10 M -13 d. [OH ] = 4.2 x 10 M pOH = - log [OH ] -13 = - log 4.2 x 10 = 13 - log 4.2 pH = 14- (13- log 4.2 ) = 1 + log 4.2 = 1.62 + -1.62 [H ] = 10 -2 = 2.4 x 10 M -
6. Calculate the OH concentration in each of following solutions in which the hydrogen ion concentrations are -8 a. 3.5 x 10 M b. 0.0065 M -13 c. 2.5 x 10 M
-5
d. 7.5 x 10 M Jawab: + OH- if H -8 a. 3.5 x 10 M pH = 8- log 3,5 pOH = 14 - pH = 14 - (8 - log 3,5) = 6 + log 3,5 = 6,5 -6,5 OH = 10 -3 b. 0.0065 M = 6.5 x 10 pH = 3- log 6,5 pOH = 14 - pH = 14 - (3 - log 6,5) = 11 + log 6,5 = 11,8 -11,8 OH = 10 -13 c. 2,5 x 10 M pH = 13- log 2,5 pOH = 14 - pH = 14 - (13 - log 2,5) = 1 + log 2,5 = 1,39 -1,39 OH = 10 -5 d. 7,5 x 10 M pH = 5- log 7,5 pOH = 14 - pH = 14 - (5 - log 7,5) = 9 + log 7,5 = 9,8 -9,8 OH = 10 7. A certain brand of beer had a hydrogen -5 ion concentration equal to 1.9 x 10 mol -1 L .What is the pH of the beer? Jawab: + pH = -log [H ] -5 = -log [1.9 x10 ] = 5-log 1.9 pH = 5 - 0.28 pH = 4.72 8. A soft drink was put on the market with [ ] = 1,4 x pH? Jawab:
mol
. What it's
+
pH = - log [H ]
-5
= - log 1,4 x 10 pH = 5 - log 1,4 9. Calculate the pH of each of the solutions in Exercises 5 and 6. Jawab: Exercise 5 a. [OH-] = x.M -3 = 2.4 x 10 -3 pOH = - log [2.4 x 10 ] = 3 - log 2.4 pH = 14 - (3 - log 2.4) = 11 + log 2.4 b. [OH-] = x.M -5 = 1.4 x 10 -5 pOH = - log [1.4 x 10 ] = 5 - log 1.4 pH = 14 - (5 - log 1.4) = 9 + log 1.4 c. [OH-] = x.M -9 = 5.6 x 10 -9 pOH = - log [5.6 x 10 ] = 9 - log 5.6 pH = 14 - (9 - log 5.6) = 5 + log 5.6 d. [OH-] = x.M -13 = 4.2 x 10 -13 pOH = - log [4.2 x 10 ] = 13 - log 4.2 pH = 14 - (13 - log 4.2) = 1 + log 4.2 Exercise 6 -8 a. [H+] = 3.5 x 10 -8 pH = - log [3.5 x 10 ] = 8 - log 3.5 -3 b. [H+] = 6.5 x 10 -3 pH = - log [6.5 x 10 ] = 3 - log 6.5 -13 c. [H+] = 2.5 x 10 -13 pH = - log [2.5 x 10 ] = 13 - log 2.5 -5 d. [H+] = 7.5 x 10 -5 pH = - log [7.5 x 10 ] = 5 - log 7.5
10. Calculate the molar concentrations of + H and OH in solution that have the following pH values. a. 3.14 b. 2.78 c. 9.25 d. 13.24 e. 5.70 Jawab: We have, a. pH = 3.14 so, pOH = 14 - 3.14 = 10.86 + - 3.14 [H ] = 10 - 10.86 [OH ] = 10 b. pH = 2.78 so, pOH = 14 - 2.78 = 11.22 + - 2.78 [H ] = 10 - 11,22 [OH ] = 10 c. pH = 9.25 so, pOH = 14 - 9.25 = 4.75 + [H ] = 10 - 9.25 - 4.75 [OH ] = 10 d. pH = 13.24 so, pOH = 14 - 13.24 = 0.76 + - 13.24 [H ] = 10 - 0.76 [OH ] = 10 e. pH = 5.70 so, pOH = 14 - 5.70 = 8.30 + - 5.70 [H ] = 10 - 8.30 [OH ] = 10 +
11. Calculate the molar concentration of H and OH in solution that have the following pOH values . a. 8.26 b. 10.25 c. 4.65 d. 6.18 e. 9.70 Jawab: a. pOH = 8.26 [ OH- ] = antilog 8.26 -8.26 = 10 -9 = 5x10 M pH = 14 - 8.26 = 5.74 [ H+ ] = antilog 5.74
-5.74
b.
c.
d.
e.
= 10 -6 = 1.819x10 M pOH = 10.25 [ OH- ] = antilog 10.25 -10.25 = 10 -11 = 5.6x10 M pH = 14 - 10.25 = 3.75 [ H+ ] = antilog 3.75 -3.75 = 10 -4 = 1.77827x10 M pOH = 4.65 [ OH- ] = antilog 4.65 -4.65 = 10 -5 = 2.2387x10 M pH = 14 - 4.65 = 9.35 [ H+ ] = antilog 9.35 -9.35 = 10 -10 = 4.467x10 M pOH = 6.18 [ OH- ] = antilog 6.18 -6.18 = 10 -7 = 6.6x10 M pH = 14 - 6.18 = 7.82 [ H+ ] = antilog 7.82 -7.82 = 10 -8 = 1.5x10 M pOH = 9.70 [ OH- ] = antilog 9.70 -9.70 = 10 -10 = 1.995x10 M pH = 14 - 9.70 = 4.3 [ H+ ] = antilog 4.3 -4.3 = 10 -5 = 5.0118x10 M
12. What is the pH of 0.010 M HCl ? Jawab: Ma + [H ]
pH
-2
= 1 x 10 mol/liter = x . Ma -2 = 1 . 1 x 10 mol/liter -2 = 1 x 10 mol/liter -2 = - log [1 x 10 ] =2
13. What is the pH of 0.0050 M solution of HNO3 ? Jawab: + [H ] = x . Ma -3 = 1 x 5 x 10 -3 = 5 x 10 + pH = - log [ H ] -3 = - log 5 x 10 = 3 - log 5 14. A sodium hydroxide solution is prepared by dissolving 6.0 g NaOH in 1.00 L of solution. What is the pOH and the pH of the solution? Jawab: D1 : m NaOH = 6.0 gram V larutan = 1.00 L Mr NaOH = 40 D2 : pOH and pH = .....? D3: [NaOH]
-
[OH ] pOH
pH
=
x
= x = 0.15 M = 0.15 M -1 = 1.5 x 10 M = - log [OH ] -1 = - log 1.5 x 10 = 1 - log 1.5 = 0.824 = 14 - 0.824 = 13.176
15. A solution was made by dissolving 0.837 g Ba(OH)2 in 100 mL final volume. What is the pOH and the pH of the solution? Jawab: D1: mass Ba(OH)2 = 0.837 g Mr Ba(OH)2 = 171 V = 100 mL D2: pOH and pH of solution = …? D3 :
Massa HCl
= mol HCl. Mr HCl -4 = 7,9. 10 . 36,5 -4 = 288,35. 10 gram -4 = 2,8835. 10 gram
18. Write the chemical equation for the ionization of each of the following weak acids in water (For any polyprotic acids , write only the equation for the first step in the ionization). a. HNO2 b. H3PO4 2c. HAsO4 d. (CH3)3 NH+ Jawab: a. + HNO2 H + NO2 +
b. H3PO4 16. A solution of Ca(OH)2 has a measured pH of 11.60. What is the molar concentration of Ca(OH)2 in the solution? Jawab: pOH = pKw - pH =14 - 11.60 =2.4 pOH = - log [OH ] -2.4 [OH ] = 10 -3 = 3.98 x 10 2+ Ca(OH)2 (aq) Ca (aq)+ 2OH (aq) [Ca(OH)2]= -3 = 1.99 x 10 So, [Ca(OH)2] = 1.99 x 10-3 17. A solution of HCl has a pH of 2.50. How many grams of HCl are there in 250 mL of this solution. Jawab: + HCl H +Cl pH HCl =2,5 + -3 [HCl] = [H ]= 3,16. 10 M Mol HCl = M HCl. V HCl -3 -1 =3,16. 10 . 2,5. 10 -4 = 7,9. 10 mol
c. HAsO4
3H + PO4 2-
3-
3-
+
AsO4 + H +
+
d. (CH3)3 NH
(CH3)3 N + H
19. For each of the acids in exercise 18, write the appropriate K a expression Jawab: + a. HNO2 H + NO2
b. H3PO4
c. HAsO4
+
3-
+
+ AsO4
3 H + PO4 2-
H
+
d. (CH3)3 NH
3-
+
(CH3)3 N + H
20. Write the chemical equation for the ionization of each of following weak bases in water. a. (CH3)3 N 3 b. AsO4 c. NO2 d. (CH3)2 N2H2 Jawab:
a. (CH3)3 N + H2O ↔ CH3 — N — H + OH CH3 32 b. AsO4 + H2O ↔HAsO4 + OH 2c. NO + H2O ↔ HNO2 + OH +
d. (CH3)2 N2H2 + H2O ↔ H — N — NH2 + OH CH3 CH3 21. For each of the bases in Exercise 20, write the appopriate K b expression. Jawab: + a. (CH3)3 N + H2O (CH3)3 NH + OH
3-
b. AsO4 +H2O H3AsO4 + H2O
-
H3AsO4+ 3OH As(OH)5
exercise 22), would produce in water as functions as a Bronsted base. Then write the expression for the K b of the conjugate base of benzoic acid. Jawab: C6H5CO2 + H2O C6H5OH + OH K b = 24. The pKa of HCN is 9.21 and that of HF is 3.17. Which is the strong Bronsted − − base CN or F ? Jawab: -10 pKa HCN=9.21→ Ka HCN= 6.17×10 -4 pKa HF = 3.17 → Ka HF = 6.76×10 −
-4
Kb CN =
= 0.16×10
−
-
c. NO2 + H2O
-
HNO2 + OH
-10
Kb F = = 0.15×10 − So, the strong Bronsted base is CN x
d. (CH3)2 N2H2 + H2O + (CH3)3 NH3 + OH
22. Benzoic acid, C6H5CO2H, is an organic acid whose sodium salt, C6H5CO2 Na, has long been used as a safe foods additive to protect beverages and many foods againts harmful yeasts and bacteria. The acid is monoprotic. Write the equation for it's Ka ! Jawab: C6H5COOH+NaOH C6H5COONa + H2O + C6H5COOH C6H5COOH + H Ka 23. Write the equation for the equilibrium that the benzoate ion, C6H5CO2 (review
25. The K a for HF is 6.8 x 10 . what is the Kb for F ? Jawab: + HF H + F Kw = Ka x Kb -14 -4 10 = 6.8x10 x Kb -11 Kb = 1.47x10 26. The barbiturate ion C4HO has Kb = 1,0 -10 x 10 . What is Ka for Barbituric acid ? Jawab: Kw = Kb x Ka -14 -10 10 = 1,0 x 10 x Ka Ka Ka
= -4 = 10
27. Hydrogen peroxide, H2O2 is a week acid -12 with Ka = 1.8 x 10 . What the value of Kb for the HO2 ion? Jawab: Kb
= =
-3
= 5.56 x 10 28. Methylamine, CH3 NH2 resambles ammonia in odor and basicity. Its K b is -4 4.4 x 10 . Calculate the K a of its conjugate acid! Jawab: K a
-2
-1
10 mol L . Calculate the Ka and pKa for periodic acid! Jawab: -2
3.8 x 10 = -2 2 ( 3.8 x 10 ) = Ka x 0.1 -3 1.44 x 10 = Ka x 0.1
=
Ka = -2 Ka = 1.44 x 10
= -11 = 2.27 x 10 29. Lactic acid, HC3H5O3, is responsible for o the sour taste of sour milk. At 25 C its -4 Ka = 1.4 x 10 . What is the Kb of its conjugate base, tha lactate ion, C3H5O3 ? Jawab: Kw = Ka xKb -14 -4 10 = 1.4 x 10 x Kb Kb Kb
= -11 = 7.14 x 10
30. Iodic acid, HIO3 has a pKa of 0.77 a. What is the formula an the Kb of its conjugate base? b. Its is conjugate base a stronger or a weaker base than the acetate ion? Jawab: a. the formula HIO3 H+ = IO3 pKa = 0.77 -1 Ka = 5,88 x 10 -14 Ka x Kb = 10 Kb = -14 = 1,7 x 10 -5 b. Ka CH3COO = 1,8 x 10 -10 Kb CH3COO = 5,5 x 10 SO, HIO3= is stronger conjugate base then asetate ion. 31. Periodic acid,HIO4,is an important oxidizing agent and a moderately strong + acid. In a 0.10 M solution , [H ] = 3.8 x
pKa = -log Ka -2 = -log (1.44 x 10 ) = 2 - log 1.44 pKa = 1.84 32. Choloacetic acid, HC2H2ClO2, is a stronger monoprotic acid than acetic acid. In a 0,10 M solution, this acid is 11 % ionized. Calculate the K a and pK a for Choloacetic acid. Jawab: α= 0,11 = 2
= (0,11) x 0,1 = 1,21 x = 3 - log 1,21 pK a = - log K a pK a = - log 1,21x 33. Ethylamine, CH3CH2 NH2, has a strong, pungent odor similar to that ammonia. Like ammonia, it is a Bronsted base. A 0.10 M solution has a pH of 11.86. Calculate the Kb and pKb for ethylamine. Jawab: M = 0.1 mol/L pH = 11.86 pOH = 14 - 11.86 = 2.14 -3 [OH-] = 7.2 x 10
-3
7.2 x 10 -5 5.184 x 10 : 0.1 = Kb -4 Kb = 5.184 x 10 34. Hidroxylamine, HONH2, like ammonia, is a Bronsted base. A 0.15 M solution has a pH of 10.12. What are Kb and pKb for Hidroxylamine? Jawab: -
[OH ] = 10
-3.88
10
=
-7.76
=
Kb = -7 Kb = 1.158 x 10 pKb = - Log Kb -7 = - Log 1.158 x 10 = 7 – Log 1.158 = 6.936 35. Refer to data in the preceding question to calculate the percentage ionization of the base in 0.15 M HONH2. Jawab:
α:
=
= 0.000878
36. What is the pH of 0.125 M pyruvic acid -3 ? It's K a is 3.2 x 10 Jawab: Ma = 0.125 mol/liter -3 K a = 3,2 x 10 +
[H ]
pH
37. What is pH of 0.15 M HN3 ? for HN3, -5 Ka = 1.8 x 10 Jawab: +
[H ] = = = -3 = 1.64 x 10 + pH = - log [ H ] -3 = -log 1.64 x 10 = 3 - log 1.64 38. What is the pH of a 1.0 M solution of hydrogen peroxide, H2O2? For this -2 solute, Ka = 1.8 x 10 Jawab: + H2O2 O2 + 2H + 2e +
[H ]
=
pH
= = 0.134 M + = - log [H ] = - log 0.134 = 0.87
39. Phenol, also known as carbolic acid, is sometimes used as a disinfectant. What are the concentrations of all of the substance in a 0.050 M solution of phenol, HC6H50? What percentage of the phenol is ionized? For this acid, K a= 1.3 x 10-10 Jawab:
= =
= -2 = 2 x 10 -2 = - log [2 x 10 ] = 2 - log 2 = 2 - 0.301 = 1.699
B R A
HC6H5O 0.05
C6H50 + -
2.55 x 10 -6 2.55 x 10-6
+
H -
2.55x10-6
0.05 - 2.55 x 10 -6 2.55 x 10 -6 -6 2.55 x 10
40. Codeine, a cough suppressant extracted from crude opium, is a weak base with a pKb of 5.79. What will be the pH of a 0.020 M solution of codeine? (Use Cod as a symbol for codeine) Jawab: pKb = - log Kb -5.79 -6 Kb = 10 = 10 [OH-]= = = -4 = 1.4 x 10 pOH = - log [OH ] -4 = - log 1.4 x 10 = 4 - log 1.4 = 4 - 0.146 = 3.854 pH = pKw - pOH = 14 - 3.854 = 10.146 So, pH of Cod = 10.146 41. Deuteroammonia, ND3, is a weak base o with a pKb of 4.96 at 25 C. What is the pH of a 0.20 M solution of this compound? Jawab: [OH]
= =
= -3 = 1,48. 10 pOH = 3-log 1,48 pH = 14 - (3-log 1,48) = 11+ log 1,48 = 11,17
42. A solution of acetic acid has a pH of 2.54. What is the concentration of acetic acid in this solution ? Jawab: + pH = - log [H ] + 2.54 = - log [H ] + -2.54 [H ] = 10 + -3 [H ] = 2.88 x 10 +
[H ]
= -3
2.88 x 10 = -3 2 -3 (2.88 x 10 ) = 1.8 x 10 x Ma -5
= 1.8 x 10 x Ma -1 3.2 x 10 = Ma 0.32 = Ma 43. Aspirin is acetylsalicyclic acid, a monoprotic acid whose K a value is 3,27 -4 x 10 . does a solution of the sodium salt of aspirin in water test acidic, basic, or neutral ? Explain Jawab: The sodium salt of aspirin is basic, because it from acetylsalyclic acid and sodium hydroxide. Weaker acid with stronger base want to produce basic salt or if sodium hydroxide dissolved in water to produce acetylsalicyclic acid and OH NaAcetylsalicyclic + H2O HAcetylsalicyclic + OH 2-
44. The K b value of the oxalate ion, C2O4 , -10 is 1.9x10 . Is a solution of K 2C2O4 acidic, basic, or neutral? Explain. Jawab: K 2C2O4 merupakan basa karena K 2C2O4 merupakan garam yang terbentuk dari basa kuat dan asam lemah sehingga garamnya bersifat basa. K 2C2O4 dapat diperoleh dari mereaksikan basa kuat yaitu KOH dan asam lemah H2C2O4. Reaksi: H2C2O4 + 2 KOH → K 2C2O4 + 2 H2O 45. Consider the following compounds and suppose that 0.5M solutions are
prepared of each : NaI, KF, (NH4)2SO4, KCN, KC2H3O2, CsNO3, and KBr. Write the formulas of those that have solutions that are a. Acidic, b. Basic, and c. Neutral. Jawab : a. Acidic is (NH4)2SO4 and CsNO3 b. Basic is KF, KCN, KC2H3O2 c. Neutral is NaI and KBr 46. Will an aqueous solution of ALCl3 turn litmus red or blue ? explain? Jawab: yes it will. based on Lewis' acid-base theory , AlCl3 is an acid . because its configuration have a vacant orbital . it means that the central atom of AlCl 3 is not octet yet . so we can assume that AlCl3 is an acidic acid based on Lewis' acid-base theory. Berdasrkan teori asam basa Lewis , senyawa AlCl3 merupakan senyawa yang bersifat asam karena atom pusatnya belum mencapai konfigurasi oktet atau bisa dikatakan masih mempunyai orbital kosong. Jadi, larutan AlCl3 dapat merubah warna kertas lakmus biru menjadi merah karena sifat keasamannya 47. Explain why the beryllium ion is a more acidic cation than the calcium ion. Jawab: in the periodic system of elements, Be is located above Ca so it is more likely to be acidic because of the periodic system of elements in one group the greater the atomic number or from top to the down it will be more likely to be basic. So, beryllium ion is a more acidic cation than the calcium ion. 48. Ammonium nitrate is commonly used in fertilizer mixtures as a source of nitrogen for plant growth. What effect,
if any, will this compound have on the acidity of the moisture in the ground? Explain. Jawab: Ammonium nitrate (NH4 NO3) + − NH4 NO3 → NH4 + NO3 + + NH4 + H2O ↔ NH3 + H3O − NO3 + H2O → If any this compound in the ground, the acidity of the moisture in the ground + will increase. There is H3O as a product from the chemical equations above. 49. Calculate the pH of 0.20 M NaCN. Jawab: [NaCN] = 0.20 M [OH-] = = = -5
pOH pH
= 1.82x10 = 5 - Log 1.82 = 9 + Log 1.82
50. Calculate the pH of 0,04 M KNO2 ? Jawab: + KNO2 → K + NO2 + K + H2O → NO2 + H2O → HNO2 + [OH ] -
[OH ] =
x [NO2]
= x -12 = 10 pOH = 12 pH = 2 51. Calculate the pH of 0.15 M CH3 NH3Cl. -4 For CH3 NH2, Kb = 4.4 x 10 Jawab: +
[H ]
=
-6 2
-10
(6.9 x 10 ) = 5.5 x 10 x = -6 = 1.846 x 10 + = - log [H ] -6 = - log (1.846 x 10 ) = 6 - log 1.846 = 5.734
pH
52. A weak base B forms the salt BHCl, + composed of the ions BH and Cl . A 0.15 M solution of the salt has a pH of 4.28. What is the value of K b for the base B? Jawab: pOH = 14 - 4.28 = 9.72 -9.27 [OH ] = 10 -10 = 1.9 x 10 -
[OH ]
=
-10
5.5 x 10 x gram NH4Br = -11 4.761 x 10 x 76 Massa NH4Br = 6.5 gram 54. The conjugate acid of a molecular base + has a hypohetical formula. BH , and has pKa of 5.00. A solution of salt of this cation, BHY, tests slightly basic. Will the conjugate acid of Y , HY, have a pKa greater than 5.00 or less than 5.00? explain Jawab: Conjugate acid is BH+ that pKa = 5 + BHOH↔ BH + OH pKa=5, Ka = 10-5 BHOH + HY ↔BHY + H2O + BHY ↔ BH + Y
-10
1.9 x 10 = -20 3.61 x 10 = K b x 0.15 -20 K b = 24.067 x 10 -19 = 2.4067 x 10 53. Calculate the number of grams of NH4Br that have to be dissolved in 1.00 o L of water at 25 C to have a solution with a pH of 5.16 ! Jawab: pH = 5.16 -5.16 [H+] = 10 -6 = 6.9 x 10 M NH4Br = = =
+
[H ] = -6 6.9 x10 =
-
[ OH ] = = = = -5 = 10 x [BHY] x 10 pOH = 5 - log [BHY] pOH = pKa HY So, pKa less than 5 55. Many drugs that are natural Bronsted bases are put into aqueous solution as their much more soluble salt with strong acids. The powerful painkiller morphine, for example, is very slightly soluble in water, but morphine nitrate is quite soluble. We may represent morphine by the symbol Mor and its + conjugate acid as H-Mor . The pKb of morphine is 6.13. What is the calculated + pH of a 0.20 M solution of H-Mor ? Jawab: pKb = 6.13 -pKb Kb = 10
-6.13
= 10 -7 Kb = 7.41 x 10 + H2O + OH
Mor + Mor 0.2 M
H0.2 M
-
[OH ]= = -7 = 7.41 x 10 pOH= 7 - log 7.41 =6.13 So, pH = 14 - 6.13 = 7.87 56. Quinine, an important drug in treating malaria, is a weak Bronsted base that we may represent as Qu. To make it more soluble in water, it is put into a solution as its conjugate acid, which we may represent as H. What is the calculate pHof a 0,15 M solution of H0
? Its pK a is 8,52 at 25 C. Jawab: D1
: [H-
] = 0,15 M, pKa = 8,52
H+ H2O D2 : pH....? D3 : pK a = 8,52 pK a = - log K a 8,52 = - log K a
Qu+H3
pH = - log = - log 2,13 x pH = 5 - log 2,13 57. Generally, under what conditions are we unable to use the initial concentration of an acid or base as though it were the equilibrium concentration in the mass action expression? Jawab: Initial concentration is unable to calculate equilibrium concentration when mole of both of components which react is same. 58. What is the percentage ionization in a 0.15 M solution of HF ? What is the pH of the solution ? Jawab: D1 : M HF = 0.15 Molar -4 Ka HF = 6.5 X 10 D2 : pH ? D3
+
: [H ] = = -3 = 9.87 x 10 + pH = - Log [H ] -3 = - Log 9.87 x 10 = 3 - Log 9.87 = 2.00568
K a = 3,02 x HM 0,15
+ H2O -
R
Qu + H3 -
-
S 0,15 -
59. What is the percentage ionization in 0.0010 M acetic acid ? What is the pH of the solution? Jawab:
α: K a =
=
[ H+ ] =
= 0.134 = -4
= 1.34x10 -4 pH = -log 1.34x10 = 4 – log 1.34 = 3.8729
3,02 x 3,02x = 2,13 x
60. What is the pH of a 1.0 x 10 solution of HCl ? Jawab:
-7
M
-7
Ma = 1.0 x 10 mol/liter Dalam hal ini berlaku ketentuan : + [H ] [OH ] = K w [Cl ] = [HCl] + [H ] = [OH ] + [Cl ] ; prinsip penetralan muatan
-7
pH = - log 1,62 x 10 = 6.79 61. The hydrogen sulfate ion HSO4 , is a moderately strong Bronsted acid with a -2 Ka of 1.0x10 . a. Write the chemical equation for the ionization of the acid and give the appropriate Ka expression. + b. What is the value of [ H ] in 0.010 M HSO4 (furnished by the salt, NaHSO4) ? Do NOT make simplifying assumptions; solve the quadratic equation. + c. What is the calculate of [H ] in 0.010 M HSO4 , obtained by using the usual simplifying assumption? d. How much error is produced by incorrectly using the simplifying assumption? Jawab: + 2a. HSO4 H + SO4
+ 2
+
because [H ] =
So, . Ka = + 2 [H ] = Ka x [HSO4 ]
-2
+
[H ] =
+
c. [H ] = = -2 = 1 x 10 d. The error has happened in using the simplifying assumption is 0 % because product of point c is equal to with point b. 62. Para-Aminobenzoic acid (PABA) is a powerful sunscreening agent whose salt were once used widely in suntanning...... The parent acid, which we may symbolize as H-Paba, is a weak acid o with a pKa of 4.92 (..... C). What is the [H+] and pH of 0.030 M solution of this acid? Jawab: 15. Given : pKa H- Paba = 4.92 [H-Paba] = 0.030 M + Asked : [H ] and pH = ...? Solution : NH2
COOH -
+
COO + H pKa 4.92 Ka
= - log Ka = - log Ka -5 = 1.2 x 10
Ka
= -5
1.2 x 10 +
b. Ka = 2[SO4 ]
-2
[H ] = 10 x 10 -2 10
=
[H ] + [H ]
= -4 = 6 x 10
pH
= - log [H ] = 4 - log 6
= 3.22
+
NH2
63. Barbituric acid, HC4H3 N2O3 (which we will abbreviate H-Bar), was discovered by the Nobel Prize-winning organic chemist Adolph von Baeyer and named after his friend, Barbara. It is the parent compound of widely sleeping drugs, the barbituretes. Its pKa is 4.01. what is the + [H ] and pH of a 0.050 M solution of HBar? Jawab: D1: pK a = 4.01 M = 0.050 mol/L +
D2 : [H ] and pH = …? D3 :
b. NaH2PO4 and Na2HPO4 (the "phosphate" buffer in side body cells) c. NH4Cl and NH3 Jawab: a. H2CO3(aq) + NaOH(aq) NaHCO3(aq) + H2O(l) Ionic equation: + 2+ 2H (aq) + CO3 (aq) + Na (aq) + OH (aq) + Na (aq) + HCO3 (aq) + H2O (l) Weak acid : H 2CO3 Conjugation base : HCO 3 b. H3PO4 (aq)+ NaOH (aq) NaH2PO4(aq) + H2O Ionic equation: + 3+ 3H (aq) + PO4 (aq) + Na (aq) + OH (aq) + Na (aq) + H2PO4 (aq) + H2O(l) H2PO4(aq) + NaOH (aq) HPO4 (aq) + H2O (l) Ionic equation: + + Na (aq) + H2PO4 (aq) + Na (aq) + + 2OH(aq) 2Na (aq) + HPO4 (aq) + H2O(l) Weak acid : H 2PO4 2Conjugation base : HPO 4 c. NH3(aq) + HCl (aq) NH4Cl (aq) Ionic equation: + NH3(aq) + H (aq) + Cl (aq) + NH4 (aq) + Cl (aq) Weak base : NH 3 + Conjugation acid : NH 4
64. Write ionic equation that illustrate how each pair of compounds can serve as a buffer pair. a. H2CO3 and NaHCO3 (the "carbonate" buffer in blood)
65. Which buffer would be better able to hold a steady pH on the addition of strong acid, buffer 1 or buffer 2? Explain. Buffer 1 is a solution containing 0.10 M NH4Cl and 1 M NH3. Buffer 2 is a solution containing 1 M NH4Cl and 0.10 M NH3. Jawab: D1: a. buffer 0.10 M NH4Cl and 1 M NH3 b. buffer1 M NH4Cl and 0.10 M NH3
d2: buffer would be better able to hold a steady pH=? D3: pH larutan a sebelum ditambah asam kuat -
[OH ] = Kb NH3. -5
= 1,8 .10 -4 = 1,8. 10 = 4- log 1,8 pOH = 3,745 pH = 10,255 pH larutan b sebelum ditambah asam kuat -
[OH ] = Kb NH3. -5
= 1,8 .10 -6 = 1,8. 10 = 6- log 1,8 pOH = 5,74 pH = 8,26 pH larutan a setelah ditambah asam kuat Misalnya asam kuat yang ditambahkan HCl 0.05 M
Perubahan pH larutan a sebelum ditambah asam kuat dan pH larutan a setelah ditambah asam kuat = (10,255 8,78) = 1,475 Perubahan pH larutan b sebelum ditambah asam kuat dan pH larutan b setelah ditambah asam kuat = (8,26 7,935) = 0,325 Kesimpulannya larutan penyangga yang pHnya cenderung hanya berubah sedikit adalah larutan b karena perubahan pHnya hanya 0,325 66. What is the pH of a solution that contains 0.15 M HC2H3O2 and 0.25 M -5 C2H3O2 ? Use Ka = 1.8 x 10 for HC2H3O2 Jawab: D1 : C2H2H3O2 H+ +C2H3O2 M HC2H3O2 : 0.15 M M C2H3O2- : 0.25 M D2 : pH ….? D3 :
-
[OH ] = Kb NH3. -5
= 1,8 .10
-5
-5
=1,8 .10 -6 = 6. 10 = 6- log 6 pOH = 5,22 pH = 8,78 pH larutan b setelah ditambah asam kuat -
1.8x10 = -5 + 1.8x10 x 0.15 = 0.25 x [H ] +
-5
1.08 x 10 pH
= [H ] + = [H ] -5 = -log 1.08 x 10 = 5-log 1.08 = 5-0.033 = 4.967
[OH ] = Kb NH3. -5
= 1,8 .10
-5
=1,8 .10 -7 = 8,6. 10 = 7- log 8,6 pOH = 6,065 pH = 7,935
67. Rework the preceding problem using the Kb for the acetate ion. ( be sure to write the poper chemical equation and equilibrium law ) Jawab: D1 : M HC2H3O2 = 0.15 M M C2H3O2 = 0.25 M -5 Ka = 1.8 x 10
D2: pH dengan menggunakan Kb dari ion asetat D3:
Atau dengan cara lain
68. By how much will the pH change if 0.050 mol of HCl is added to 1.00 L off the buffer in Exercise 66. Jawab: [H+] = Ka x -5
= 1.8 x 10 x -5 =1.081 x 10 pH= -log [H+] = - log 1.081 x 10-5 = 5 - log 1.081 = 4.9664 Perubahan pH= 4.9666 - 4.9664 = 0,0002 Perubahan pH sangat kecil karena jumlah HCL yang di tambahkan sangat sedikit sedangkan volume buffernya besar. 69. By how much will the pH change if 50.0 mL of 0.10 M NaOH is added to 500mL of the buffer in Exercise 66. Jawab: D1 : 500 mL of Buffer that contain 0,15 M HC2H3O2 and 0,25 M C2H3O2 ? Ka = 1,8 x 10-5 for HC2H3O2 D2 : How much will pH Change ? D3: HC2H3O2 + NaOH
m = 75 mmol 5 mmol r = 5 mmol 5 mmol _ s = 70 mmol 0
C2H3O2 Na + H2O
125 mmol 5 mmol 130 mmol
+
70. A buffer is prepared containg 0.25 M NH3 and 0.14 M NH4+ a. calculate the pH of the buffer using the Kb for NH3 b. calculate the pH of the buffer using + the Ka for NH4 Jawab: + NH3 (aq) + H2O(l) NH4 (aq) + OH -5 a. misal Kb = 1.8 x 10 dan volume larutan dianggap sama , maka -
[OH ]=Kb . = Kb . = Kb . -5 = 1.8 x 10 . 1.7857143 -5 = 3.21428574 x 10 pOH = 5 - log 2.21428574 pOH = 4.492915518 pH = 9.507084482 -5 b. misal Ka = 10 +
[H ] = Ka
-5
= 0.56x10 -6 = 5.6x10 pH = 6 - log 5.6 pH = 5.25 If added 0.020mol HCl
-5
= 10
-5
= 10 -5 = 10 x 0.56 -6 = 5.6 x 10 pH = 6 - log 5.6 71. By how much will the pH change if 0.020 mL of HCl is added to 1.00 L of the buffer in Exercise 70? Jawab: D1 : Buffer + 0.25 NH3 and 0.14 NH4 1.00 L D2 : the change of pH if 0.020 mol HCl is added to 1.00 L.? D3 : the mol of base = 0.25 M x 1.00 L = 0.25 mol The mol of conjugate acid = 0.14 M x 1.00 L = 0.14 mol + * I f using K of NH 3 b
+
[H ] = K a -5
= 10 x -5 = 1.43x10 pH = 5 - log 1.43 pH = 4.84 Change of pH = 5.25 – 4.84 = 0.41 72. By how much will the pH change if 75 ml of 0.10 M KOH is added to 200 ml of the buffer in exercize 70? Jawab: pH buffer
−
[OH ] = Kb ×
-
[OH ] = K b
-5
-5
= 1.8x10 x -5 = 3.21x10 -5 pOH = -log 3.21x10 = 4.5 pH = 14 - 4.5 =9.5 If added 0.020 mol HCl + + NH3 (aq) + H (aq) NH4 m 0.25mol 0.020mol 0.14 mol r - 0.020mol 0.020mol +0.02mol s 0.23mol 0.16mol
= 1×10 × -5 = 1.79×10 -5 pOH = −log 1.79×10 = 5 - log 1.79 = 5 – 0.25 = 4.75 pH = 14 – 4.75 = 9.25 Added by 75 ml 0.10 KOH NH4
-
[OH ]
-5
= 1.8x10 x -5 = 2.6x10 pOH = 5-log 2.6 = 5-0.41 = 4.59 pH = 14 - 4.59 = 9.41 the change of pH = 9.5 - 9.41 = 0.09
m r s
+
28 mmol 7.5 mmol 7.5 mmol 7.5 mmol 20.5 mmol −
−
-5
+
+
-5
= 10 x
50 mmol 7.5 mmol 57.5 mmol
[OH ] = Kb ×
* I f using K of NH 4 a
[H ] = K a
− (aq)+OH (aq)→NH3(aq)+H2O(l)
pOH
= 1×10 × -5 = 2.8×10 -5 = −log 2.8×10 = 5 – log 2.8 = 5 - 0.45
= 4.55 pH = 14 - 4.55 = 9.45 So, pH is change from 9.25 to 9.45 73. How many grams of sodium acetat, NaC2H3O2, would have to be added to 1.0 L of 0.15 M acetic acid (pK a 4.74) to make the solution a buffer for pH 5.00? Jawab: Jawab: + [H ] = -5
10
=
X
= = 0.0833
Mass of CH3COOH = mol x Mr = 0.0833 x 60 = 5 gram
75. What mole ratio of NH4Cl to NH3 would buffer a solution at pH 9.25? Jawab: pOH = 4.75 -
[OH ]
= Kb -5
1.77 x 10
= = 76. How many grams of ammonium choride would have to be dissolved in 500 mL of 0.20 M NH3 to prepare a solution buffered at pH 10.00? Jawab: D1 :500 mL of 0.20 M NH3 pH buffer = 10.00 D2: m NH4Cl = ...? -4 D3: pH = 10, pOH = 4, [OH ] = 10 -
74. How many grams of sodium formate, NaCHO2, would have to be added to 1.0 L of 0.12 M formic acid (Pk a 3.74) to make the solution a buffer for pH 3.80 ? Jawab: CHO2H 1.0 L 0.12 M = 0.12 mol Mr NaC2HO2 = 68 -4 pKa = 3.74 Ka = 1.8 x 10 + -4 pH = 3.80 [H ] = 1.58 x 10 kasus buffer asam [H+]
= Ka
1.58 x 10
-4
= 1.8 x 10
1.58 x 10 X
-4
= = 0,14 mol
n
-4
=
0.14 = massa NaC2HO2 = 9.3 gram
-5
= 1.8 x 10
[OH ] -14
= Kb -14
10 -14 10 n AK
= 1.8 x 10 . = 0.018 x n AK -3 = 5.556 x 10
n AK
= -3
5.556 x 10 = m = 0.297 gram 77. How many grams of ammonium chloride have to be dissolved into 125 mL of 0.10 M NH3 to make it a buffer with a pH of 9.15 ? Jawab: D1 : V NH3 = 125 mL M NH3= 0.10 mol/L pH = 9.15 -4.85 Kb = 1.8 x 10 D2: massa of NH4Cl . . .? D3: Mol NH3 = 12.5 mmol pH = 9.15 pOH = 4.85
[OH-]
= Kb -5
79. How many milliliters of 0.15 M HCl would have to be added to 100 mL of the buffer described in exercise 78 to make the pH decrease by 0.05 pH unit? How many milliliters of the same HCl solution would, if added to 100 mL of pure water, make the pH decrease by 0.05 pH unit? Jawab:
-5
1.4 x 10 = 1.8 x 10 -4 -5 2.25 x 10 = n asam konj. x 1.4 x 10 n as.konj = 16.07 mmol n as.konj = massa NH4Cl = 16.07 x 53.5 = 859.75 mg = 0.86 gram 78. Suppose 25.00 mL of 0.100 M HCl is added to an acetate buffer prepared by dissolving 0.100 mol of acetic acidand 0.110 of sodium acetate in 500 mL of solution. What are the initial and final pH value? what would be the pH if the same amount of HCl solution were added to 125 mL of pure water? Jawab: Awal : +
[ H ] = Ka -5
= 1,8 x 10 x -5 = 1,64 x 10 pH = 5 - log 1,64 = 4, 79 + [H ] =
Ka
x
-5
= 1,8 x 10 x
+
a) [H ]
= -5
4.45 x 10 = -6 -5 -6 4.89 x 10 - 4.45 x 10 a = 1.8 x 10 + 1.8 -5 x 10 a -6 -5 4.89 x 10 - 1.8 x 10 = 1.8 x 10 a + 4.45 -5 x 10 a -6 -5 3.09 x 10 = 6.25 x 10 a a = a = 0.049 mol HCl = 0.049 M.V = 0.049 0.15 . V = 0.049 V= V= 0.33 L V= 330 mL b) Jumlah HCl yang ditambahkan adalah sama yaitu 330 mL,Buffer ditambah dengan air (pengenceran) pH larutan tetap karena penentu pH buffer adalah jumlah mol bukan konsentrasi buffer.
-5
= 1,8x 10 x -5 = 1,72 x 10 pH = 5 - log 1,72 = 4,76 Final: Pengenceran dengan penambahan air pada buffer, pH akan tetap karena penentu pH buffer adalah jumlah mol bukan konsentrasi buffer
80. What can make the titrated solution at the equivalence point in an acid-base titration have a pH not equal to 7,00 ? Ho w does this possibility affect the choice of an indicator ? Jawab: Apabila dalam titrasi asam-basa, titk ekuivalen dapat tercapai pada pH tidak sama dengan 7 berarti salah satu larutan asam atau basanya bersifat lemah.
Pemilihan indikator sangat mempengaruhi proses titrasi karena untuk mengetahui titik akhir titrasi dengan ditandai perubahan warna, sesuai dengan rentang pH yang diperkirakan sehingga pemilihan indikator yang digunakan tidak salah. 81. Explain why ethyl red is a better indicator than phenolphtalein in the titration of dilute ammonia by dilute hydrochloric acid? Jawab: Methyl red is a better indicator than phenolphtalein in the titration of dilute ammonia by dilute hydrochloric acid because the result of ammonia and hydrochloric acid is a solution that has pH < 7 ( influenced by hydrochloric acid as a strong acid and ammonia is a weak base ). Which is pH range of methyl red is 4.4 -6.2 and pH range of phenolphtalein is 8.3 -10.0 82. What is a good indicator for titrating potassium hydroxide with hydrobromic acid? Explain. Jawab: For titrating potassium hydroxide with hydrobromic acid we use metilred indicator as a good indicator. Titrating potassium hydroxide with hydrobromic acid is example of titration strong base and weak acid. The equivalen point is occur in value of pH smaller than 7, so we must use indicator that have trayek of pH under 7, for example indicator metilred that have trayek of pH from 4.8 until 6. 83. In the titration of an acid with base,what condition concerning the quantities of reactans ought to be true at the equivalence point? Jawab: the quantities of reactans ought to be true at the equivalence point when the
mols equivalence of acis as same as the mols equivalence of base. 84. When 50 mL of 0.10 M formic acid is titrated with 0.10 M sodium hydroxide, what is the pH at the equivalence point? (Be sure to take into account the change in volume during the titration). What is a good indicator for this titration? Jawab: Va = 50 ml = 0.05 L ; Ma = 0.1 mol/liter ; na = 0.05 x 0.1 = 0.005 mol M b = 0.1 mol/liter Va x Ma = V b x M b 0.05 x 0.1 = V b x 0.1 V b = 0.05 L n b = 0.05 x 0.01 = 0.005 mol Vtotal = 0.05 + 0.05 = 0.1 -4 K a = 1.8 x 10 M : 0.005 mol R : 0.005 mol mol S: mol
0.005 mol 0.005 mol
0.005 -
-
0.005
-
[OH ] = = = -6 = 1.67 x 10 pOH = 6 - log 1.67 pH = 14 - 6 + log 1.67 = 8 + log 1.67 = 8 + 0.223 = 8.223 Because it is on route pH 5.2 to 6.8, the indicator used is bromine cresol purple. 85. When 25 mL of 0.10 M aqueous ammonia is titrated with 0.10 M hydrobromic acid, what is the pH at the equivalence point? What is a good indicator?
Jawab: D1: V NH3 = 25 mL M NH3 = 0.10 mol/L -5 Kb NH3 = 1.8 x 10 M HBr = 0.10 mol/L
D2 : what is pH at equivalence point and a good indicator = …? D3: At equivalence point means that the number of acid moles equal to the moles of base. NH3 + HBR NH4Br m r s
2.5 mmol 2.5 mmol 2.5 mmol 2.5 mmol -
2.5 mmol 2.5 mmol
a. 0 mL b. 10.00 mL c. 24.90 mL d. 24.99 mL e. 25.00 mL f. 25.01 mL g. 25.10 mL h. 26.00 mL i. 50.00 mL Jawab: D1 : [HCl] = 0.1000 M [NaOH] = 0.1000 M V HCl = 25 mL D2 : pH and the titration curve = ...? Solution : a. V NaOH = 0 mL HCl + NaOH NaCl +H2O M 2.5 0 R 0 0 0 0 S 2.5 0 0 [HCl] +
[H ] pH
= = 0.1 = 1 x 0.1 = 0.1 + = - log [H ] = - log 0.1 =1
b.
V NaOH = 10.00 mL HCl + NaOH NaCl + H2O M 2.5 1 R 1 1 1 1 S 1.5 1 1
[HCl] 86. For the titratin of 25.00 mL of 0.1000 M HCl with 0.1000 M NaOH, calculate the pH of the resulting solution after each of the following quantities of base has been added to the original solution (you must take into account the change in total volume). Construct a graph showing the titration curve for this experiment.
+
[H ] pH
= = 0.043 = 1 x 0.043 = 0.043 + = - log [H ] = - log 0.043 = 1.37
-5
c.
V NaOH = 24.90 mL HCl + NaOH NaCl + H2O M 2.5 2.49 R 2.49 2.49 2.49 2.49 S 0.01 2.49 2.49
[HCl] +
[H ] pH
= -4 = 2.0 x 10 -4 = 1 x (2.0 x 10 ) -4 = 2.0 x 10 + = - log [H ] -4 = - log 2.0 x 10 = 3.698
pOH
pH
g.
V NaOH = 25.10 mL HCl + NaOH NaCl + H2O M 2.5 2.51 R 2.5 2.5 2.5 2.5 S 0.01 2.5 2.5
[NaOH] d. V NaOH = 24.99 mL HCl + NaOH NaCl + H2O M 2.5 2.499 R 2.499 2.499 2.499 2.499 -3 S 1x10 2.499 2.499
-
[OH ] pOH
[HCl]
M R S
= -5 = 2.0 x 10 + -5 [H ] = 1 x (2.0x10 ) -5 = 2.0 x 10 + pH = - log [H ] -5 = - log 2.0 x 10 = 4.7 e. V NaOH = 25.00 mL HCl + NaOH NaCl + H2O 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5
pH
V NaOH = 26.00 mL HCl + NaOH NaCl + H2O M 2.5 2.6 R 2.5 2.5 2.5 2.5 S 0.1 2.5 2.5
[NaOH] -
Titration in equivalent point pH = 7 (neutral)
pOH
f. M R S
[NaOH] -
[OH ]
= -5 = 1.9996 x 10 -5 = 1 x (1.9996 x 10 )
= -4 = 1.996 x 10 -4 = 1 x (1.996 x 10 ) -4 = 1.996 x 10 = - log [OH ] -4 = - log 1.996 x 10 = 3.7 = 14 - 3.7 = 10.3
h.
[OH ]
V NaOH = 25.01 mL HCl + NaOH NaCl + H2O 2.5 2.501 2.5 2.5 2.5 2.5 -3 1x10 2.5 2.5
= 1.9996 x 10 = - log [OH ] -5 = - log 1.9996 x 10 = 4.699 = 14 - 4.699 = 9.3
pH
i.
= -3 = 1.96 x 10 -3 = 1 x (1.96 x 10 ) -3 = 1.96 x 10 = - log [OH ] -3 = - log 1.96 x 10 = 2.7 = 14 - 2.7 = 11.3
V NaOH = 50.00 mL HCl + NaOH NaCl + H2O M 2.5 5 R 2.5 2.5 2.5 2.5 S 2.5 2.5 2.5
[NaOH] -
[OH ] pOH
pH
= = 0.033 = 1 x (0.033) = 0.033 = - log [OH ] = - log 0.033 = 1.477 = 14 - 1.477 = 12.523
b. After 10.00 mL of the base has been added means that we calculate pH of acid buffer.
CH3COOH+ NaOH B R A
2.5 mmol 1 mmol 1 mmol 1 mmol 1.5 mmol -
CH3COONa + H2O 1 mmol 1 mmol
1 mmol 1 mmol
c. At a half of the HC2H3O2 means that the number of acid moles equal to a half the moles of base. 87. For the titration of 25.00 mL of 0.1000 M acetic acid with 0.1000 M NaOH, calculate the pH: a. Before the addition of any NaOH solution, b. After 10.00 mL of the base has been added, c. After half of the HC2H302 has been neutralized, and d. At the equivalence point. Jawab: a. Before the addition of any NaOH solution, it means calculate pH of a weak acid.
CH3COOH + NaOH CH3COONa+H2O B 2.5 mmol 1.25 mmol R 1.25 mmol 1.25 mmol 1.25mmol 1.25mmol A 1.25 mmol 1.25 mmol 1.25mmol
log1.8 = 4.744 d. At equivalence point means that the number of acid moles equal to the moles of base. B R A
CH3COOH + NaOH CH3COONa +H2O 2.5 mmol 2.5 mmol 2.5 mmol 2.5 mmol 2.5 mmol 2.5mmol 2.5 mmol 2.5mmol
So, pH of NH3 before the addition of any HCl solution are 11 b. Moles of NH3 = n x M = 25 x 0.1 = 2.5 mmol Moles of HCl = n x M = 10 x 0.1 = 1 mmol NH3(aq) + HCl(aq) NH4Cl(aq) Before: 2,5 1 React: 1 1 1 After : 1.5mmol 1 mmol -
[OH ] = Kb x -5
88. For the titration of 25.00 mL of 0.1000 M ammonia with 0.1000 M HCl, calculate the pH a. before the addition of any HCl solution, b. after 10.00 mL of the acid has been added, c. after half of the NH3 has been neutralized, and d. at the equivalence point Jawab:
= 10 x -5 = 1.5 x 10 pOH = - log [OH ] -5 = - log 1.5 x 10 = 5 - log 1.5 = 5 - 0.176 = 4.824 pH = pKw - pOH = 14 - 4.824 = 9.176 So, pH after 10.00 mL of HCl has been added were 9.176 c. pH after half the NH3 has been neutralized….? NH3(aq) + HCl(aq) NH4Cl(aq) Before: 2,5 1.25 React: 1.25 1.25 1.25 After : 1.25 mmol 1.25mmol
-
a. [OH ] = = = -3 = 10 pOH = - log [OH ] -3 = - log 10 =3 pH = pKw - pOH = 14 - 3 = 11
-
[OH ] = Kb x -5
= 10 x -5 = 10 pOH = - log [OH ] -5 = - log 10 =5 pH = pKw - pOH = 14 - 5 =9
So, pH after half of the NH3 has been neutralized were 9 d. pH at the equivalence point….? NH3(aq) + HCl(aq) NH4Cl(aq) Before: 2,5 2.5 React: 2.5 2.5 2.5 After : 2.5 mmol Looking for volume total: Moles of NH3 = moles of HCl 25 x 0.1 = V x 0.1 V = 25 mL [H+] = = = -5 = 0,7 x 10 + pH = - log [H ] -5 = - log 0,7 x 10 = 5 - log 0.7 = 5 - (-0.15) = 5.15 So, pH at the equivalence point are 5.15
