The amount of lactic acid, `HC_(3)H_(5)O_(3)`, produced in a sample of muscle tissue was analysed by reaction with hydroxide ion. Hydroxide ion was produced in the sample mixture by electrolysis. The cathode reaction was,
`2H_(2)O(l)+2e^(-) rarr H_(2)(g)+2OH^(-) (aq.)`
Hydroxide ion reacts with lactic acid as soon as it is produced. The end point of the reaction is detected with an acid-base indicator. It required 115 seconds for a current of 15.6 mA to reach end point. How many grams of lactic acid (a monoprotic acid) were present in the sample ?

Answer

`1.674 xx 10^(-3) g`

Explanation

No. of moles of lactic acid = No. of moles of `OH^(-)` used
=No. of faraday used in electrolysis
Number of faradat used `=(I xx t)/(96500) =(15.6xx10^(-3) xx115)/(96500)`
`= 1.86 xx 10^(-5)`
Mass of lactic acid = Number of moles `xx` Molecular mass
`=1.86 xx 10^(-5) xx 90 =1.674 xx 10^(-3) g`