# Question bd7ca

Feb 9, 2015

The pH of the $\text{NaOH/CH"_3"COOH}$ buffer will be equal to $\text{4.75}$.

The first thing you have to do is determine how many moles of $N a O H$ and of $C {H}_{3} C O O H$ you'll have in solution

${n}_{N a O H} = 50 \cdot {10}^{- 3} \text{L" * "0.1 M" = "0.005 moles}$, and

${n}_{C {H}_{3} C O O H} = 100 \cdot {10}^{- 3} \text{L" * "0.1 M" = "0.01 moles}$

Since $N a O H$ is a strong base, it will react with the acetic acid completely. Use the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart) to determine how many moles of each species will be left after the reaction

$N a O {H}_{\left(a q\right)} + C {H}_{3} C O O {H}_{\left(a q\right)} \to C {H}_{3} C O O N {a}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$
I......0.005.................0.01.................................0
C......(-0.005)............(-0.005)........................(+0.005)
E........0.......................0.005..........................0.005

The $N a O H$ will be completely consumed in the reaction; the concentrations of $C {H}_{3} C O O H$ and $C {H}_{3} C O O N a$ wil be

C_(CH_3COOH) = n/V_("total") = ("0.005 moles")/((100 + 50) * 10^(-3)"L") = "0.033 M", and

C_(CH_3COONa) = ("0.005 moles")/(150 * 10^(-3)"L") = "0.033 M"

Since you're dealing with a buffer, use the Henderson-Hasselbalch equation to determine the pH of the solution

$p {H}_{\text{solution}} = p K a + \log \left(\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]}\right)$

pH_("solution") = 4.75 + log(("0.033 M")/("0.033 M")) = 4.75 + log(1) #

$p {H}_{\text{solution}} = 4.75 + 0 = 4.75$