Date: August 23, 2018 Group No: 7 (Guangco, S.M; Hernandez, Hernandez, F.M; Jimenez, J.) EXPERIMENT NO. 5 ASSAY OF SODIUM SODIUM BICARBONATE BICARBONATE TABLETS I.
II.
OBJECTIVES To determine the active amount of sodium bicarbonate tablets and identify if the sample complies with USP specification (USP XIX NF XIV) DATA AND RESULTS TRIAL 1 2 3 4 Weight of sample (g) 0.6991 0.6888 0.6944 0.7014 Volume of 1N HCl (mL) 8.70 8.60 8.70 8.50 % NaHCO3 116.05% 116.43% 116.83% 113.01%
Average Normality (N): 1.11N Normality Factor (Ave N/Theoretical N): 1.11N Average % NaHCO NaHCO3: 115.58% III.
OBSERVATION In the direct acidimetric titration of sodium bicarbonate with 1.11N HCl, the first duplicate trials both exceeded the pale pink coloration denoted by the methyl red indicator, despite the attempt to dispel the carbonic acid from the flask. For the succeeding trials, a fainter shade of pink was was observed up until the fourth (and last) trial where the faint pink coloration was achieved. As it may also also be possible to interfere interfere with the analysis, analysis, it is possible that further error could have been prevented if pre-boiled, CO2 – free water was used in the preparation, instead of merely using the distilled water provided by the laboratory staff. (cite reference)
IV.
CONCLUSION Based on the USP 19 NF 14 procedures on the assay of sodium bicarbonate tablets, the average %NaHCO 3 determined from the four trials was 115.58%, and has failed to meet the specifications for NaHCO 3, USP which was not less than 99% but not more than 105%
V.
ANSWERS TO QUESTIONS 1.
If exactly 3g of sodium bicarbonate is dissolved in 26 mL of water, what is the normality of the solution?
) = ( 3.00 26 = (0.08401 ) 3.00 = 0.08401 26.00 = . . ..
Date: August 23, 2018 Group No: 7 (Guangco, S.M; Hernandez, Hernandez, F.M; Jimenez, J.)
2.
If a 0.27g sample of sodium bicarbonate (96.5% NaHCO3) is titrated with 0.9265N 0 .9265N HCl, what volume of the acid is required to produce and endpoint?
100 %= %= 100 0.08401 100 96.5%= 0.9265 0.27 96.5% 0.27 = 0.9265 0.08401 100 96.5% 0.27 0.9265 0.08401 100 = 0.9265 0.08401 100 0.9265 0.08401 100 0.27 = 0.926596.5% 0.08401 100 = . . / / .. 3.
Calculate the following titer values of 1N HCl a. KHCO3
) = ( 1 1 1 = (0.10012 )
= 100. 100.12 12 b. K2CO3
) = ( 1 1 1 = (0.069105 ) = 69.1 69.11 1 c.
CaCO3
) = ( ) 1 1 1 = (0.050045 = 50.0 50.055 VI. REFERENCES Knevel, A. M., & DiGangi, F. E. (1977). Jenkins' Quantitative Pharmaceutical Chemistry. McGraw Chemistry. McGraw - Hill Inc.
