"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:
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