ElectrochemistryHard
Question
Lactic acid, HC3H5O3, produced in 1 g sample of muscle tissue was titrated using phenolphthalein as indicator against OH− ions which were obtained by the electrolysis of water. As soon as OH− ions are produced, they react with lactic acid and at complete neutralization, immediately a pink colour is noticed. If electrolysis was made for 1158 s using 50.0 mA current to reach the end point, what was the percentage of lactic acid in muscle tissue?
Options
A.5.4%
B.2.7%
C.10.8%
D.0.054%
Solution
$n_{eq}HC_{3}H_{5}O_{3} = n_{OH^{-}} = \frac{Q}{F}$
$\text{Or, }\frac{w}{90} \times 1 = \frac{50 \times 10^{- 3} \times 1158}{96500} \Rightarrow w = 0.054\text{ gm} $$$\therefore\text{\% of lactice acid = }\frac{0.054}{1} \times 100 = 5.4\%$$
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