May 15, 2015

The pH of the solution will be equal to 12.22.

So, you're titrating acetic acid, a weak acid, with sodium hydroxide, a strong base. The balanced chemical equation for this neutralization reaction is

$C {H}_{3} C O O {H}_{\left(a q\right)} + O {H}_{\left(a q\right)}^{-} \to C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)}$

SIDE NOTE The sodium cations will be spectator ions, so I didn't include them in the equation.

Notice the $1 : 1$ mole ratio that exists between acetic acid and sodium hydroxide. This means that 1 mole of the former will react with 1 mole of the latter to produce sodium acetate and water.

Determine how many moles of each species you have

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{C {H}_{3} C O O H} = \text{0.100 M" * 25 * 10^(-3)"L" = "0.0025 moles}$

${n}_{N a O H} = \text{0.100 M" * 35 * 10^(-3)"L" = "0.0035 moles}$

Notice that you have more moles of sodium hydroxide than of acetic acid. This means that the acetic acid will be consumed completely, and you'll only be left with sodium hydroxide in solution.

${n}_{C {H}_{3} C O O H} = 0$

${n}_{N a O H} = 0.0035 - 0.0025 = \text{0.0010 moles}$ $N a O H$

The total volume of the solution will be

${V}_{\text{sol" = V_"acetic acid" + V_"sodium hydroxide}}$

${V}_{\text{sol" = 25 + 35 = "60. mL}}$

This means that the concentration of the hydroxide ions will be equal to

$\left[O {H}^{-}\right] = \text{0.0010 moles"/(60. * 10^(-3)"L") = "0.01667 M}$

The pOH of the solution will be

$p O H = - \log \left(\left[O {H}^{-}\right]\right) = - \log \left(0.01667\right) = \text{1.78}$

Therefore, the pH of the solution will be

$p {H}_{\text{sol}} = 14 - p O H = 14 - 1.78 = \textcolor{g r e e n}{12.22}$