Prof. S. Sudhakar - Two Marks Questions and Answers for B.E Aeronautical Engineering Students.
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Solubility Solubility Questions and answers Question 1 Show, by calculation, that a precipitate forms when 50.0 mL of 0.100 mol L –1 KClaq KClaq!! and 50.0 mL –1 of 0.0"00 mol L #$%&'aq! aq! are mi(ed. K s #$Cl! ) 1.5* + 10 –10 After mixing volume = 100 100 mL [Ag+ ] = 0.020 2 = 0.01 mol L –1 [Cl ] – ] = 0.100 2 = 0.05 mol L–1 IP = [Ag+ ][Cl– ] ] = 0.01 ⋅ 0.05 = 5 ⋅ 10–4 IP > ! !o "re#i"it$te form!.
Question " # 50.0 50.0 mL sample sample of a saturated aueous aueous solution solution of lead bromide, bromide, -br ", was e/aporated to dryness. 0."" $ of solid -br " was obtained. a! a!
i! i!
rit rite e the the eua euati tion on for for the the eu euil ilib ibri rium um pre prese sent nt in in a sat satur urat ated ed sol solut utio ion n of lea lead d brom bromid ide. e.
ii! ii!
Compl omplet ete e the the e(p e(pre ress ssio ion n for for K s-br "!.
%$ % $&
(b) (b)
%i&
P'(r2 %!&
2+
P' P'
%$)& + 2(r *%$)&
Calcul lculat ate e the the val value of K s(PbBr ). 2 2
– M (PbBr 2 ) = 367 g mol 2 %'& n%P'(r 2 & in 50 mL = 0.422 ,,= 1.15 ⋅ 10 mol n%P'(r 2 & in 1 L = 1.15 ⋅ 10 ⋅ 20 = 2.0 ⋅ 10 mol !olu'ilit%!& of P'(r 2 = 2.0 ⋅ 10 mol L or /ire#tl n%P'(r 2 & in 1 L = .44 ,,= 2.0 ⋅ 10 mol [P' ] =2.0 ⋅ 10 mol L [(r ]= 4.,0 10 mol L ! = [P' ][(r ]2 = 2.0 ⋅ 10 ⋅ %4.,0 ⋅ 10 &2 = 4.- ⋅ 10 r 4! = 4%2.0 ⋅ 10 & –
–
–2 –2
–2
2+
–
2+
–2
*1
–2
*1
–
–2
–2
–5
–2
!= 4.- ⋅ 10 % !.f.& –5
*1
%ii&
! = [P' ][(r ] ! ex"re!!ion 2+
– 2
Question ' (a) (a)
Desc Descri ribe be what what is mean meantt by by the the term term ‘solubility’.
%$& 3olu'ilit i! te $mount of !u'!t$n#e t$t ill /i!!olve in $ given volume to form $ !$tur$te/ !olution %$t t$t tem"er$ture&.
(b) (b)
The The solu solubi bili lity ty prod produc uct, t, s, of AgCl has a alue of !."# × !$ %!$ at &"'C and this alue increases to &.!" × !$ % at !$$'C. *plain why s is higher at !$$'C. +nclude reference to the releant euilibrium euation in your answer.
%'& ! i! $n e)uili'rium e)uili'rium #on!t$nt for te re$#tion AgCl%!& Ag+ %$)& + Cl– %$)& AgCl%!& 6 Ag+%$)& + Cl–%$)& $n/ ! = [Ag+ ][Cl ] – ] 7ore !oli/ /i!!olve! en te tem"er$ture i! in#re$!e/ $! e)uili'rium !ift! in te en/otermi# /ire#tion8 i# me$n! it !ift! to te rigt $n/ i n#re$!e! te #on#entr$tion of ion! in !olution. 9i! in#re$!e! !. An!er re#ogni!e! t$t in#re$!e/ ! v$lue me$n! in#re$!e/ !olu'ilit or #on#entr$tion of ion!
The chloride ion concentration in sea water can be determined by titrating a sample with aueous siler nitrate (Ag-/) using potassium chromate (0 &Cr1) as the indicator. As the siler nitrate is added, a precipitate of siler chloride, (AgCl) forms. 2hen most of the AgCl has precipitated, the Ag3($)) concentration becomes high enough for a red precipitate of Ag &Cr1 to form. (c) (c)
4how 4how that that the the sol solub ubil ilit ity y of of Ag Ag&Cr1 in pure water at &" oC is higher than that of AgCl.
(d) +f the the concen concentrat tration ion of of chroma chromate te ions ions is #./$ #./$ 6 !$ !$ %/ mol 7 %! at the point when the Ag&Cr1 starts to precipitate, calculate the concentration of Ag3 ions in the solution. (d)