# Question b6a38

Feb 3, 2015

Start by figuring out what the concentrations the components of the buffer will have in the total solution.

The total volume of the buffer will be

V_("total") = "250 mL" + "250 mL" = "500 mL"

The number of moles of acetic acid you'll use for the buffer is

$C = \frac{n}{V} \implies n = C \cdot V = 250 \cdot {10}^{- 3} \text{L" * "0.0100 M" = 2.5 * 10^(-3)"moles}$

This means that the concentration of acetic acid in the buffer will be

C_("buffer") = n/V_("total") = (2.5 * 10^(-3)"moles")/(500 * 10^(-3)"L") = "0.005 M"

Now do the same for sodium acetate

$n = C \cdot V = 250 \cdot {10}^{- 3} \text{L" * "0.100 M" = 25 * 10^(-3)"moles}$

This means that the concentration of sodium acetate in the buffer will be

C_("buffer") = n/V_("total") = (25 * 10^(-3)"moles")/(500 * 10^(-3)"L") = "0.05 M"#

All you have to do now is use the Henderson-Hasselbalch equation to determine the pH of the buffer

$\left[p H\right] \left(h \texttt{p} : / s o c r a t i c . \mathmr{and} \frac{g}{c} h e m i s t r \frac{y}{a} c i \mathrm{ds} - \mathmr{and} - b a s e \frac{s}{t} h e - p h - c o n c e p t\right) = p K a + \log \left(\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]}\right)$

$p H = 4.76 + \log \left(\left(\text{0.05 M")/("0.005 M}\right)\right) = 4.76 + 1 = 5.76$