Question

The pH of the solution is 6 and its volume is 60 gal. How much of 12N NaOH should be added to adjust the pH of the solution to 6.5.
Also, Convert 0.1N HCl to molarity.

Answer 1

$pH(before \space adding \space NaOH) = -log[H_3O^+]=6$

$[H_3O^+](initial)=1\cdot10^{-6}M$

$pH(after \space adding \space NaOH)=-log[H_3O^+]=6.5$

$[H_3O^+](after \space adding \space NaOH)= 1\cdot10^{-6.5}M$

$V (initial \space solution)=60gal\cdot\frac{3.785L}{1gal}=227.125L$

$[H_3O^+](after\space adding \space NaOH)=\frac{V(initial \space solution)\cdot[H_3O+](initial)- V(NaOH)(added)\cdot[NaOH]}{V(initial \space solution) + V(NaOH)(added)}$

$1\cdot10^{-6.5}=\frac{227.125\space \cdot \space 10^{-6} - V(NaOH)added \space \cdot \space12}{227.125+V(NaOH)added}$

$12.9 \cdot 10^{-6}L = 12.9\mu L$ - volume of NaOH to be added to adjust pH to 6.5.

Normality of 0.1N HCl in the reaction with NaOH can be found by multiplying the normal concentration by the number of equivalents per mole:

$$M=N \cdot eq \space = 0.1 \cdot 1 = 0.1M$$