# Question 6c230

Feb 3, 2015

The total volume of the buffer represents the sum of the volumes of the acetic acid and sodium acetate solutions. I'll use ${V}_{1}$ to represent the volume of acetic acid and ${V}_{2}$ the volume of sodium acetate. Therefore,

${V}_{1} + {V}_{2} = 200 \cdot {10}^{- 3} = 0.2$ (1)

This will represent your first equation (notice that I've used liters instead of mililiters).

Since you're dealing with a buffer, use the Henderson-Hasselbalch equation to try and find another relationship between ${V}_{1}$ and ${V}_{2}$.

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

Plug in the values you have for pH and $\text{pKa}$

$3.95 = 4.76 + \log \left(\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]}\right)$, or

$\log \left(\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]}\right) = - 0.81 \implies \frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]} = 0.155$ (2)

Now, the concentration of sodium acetate can be written as

$\left[C {H}_{3} C O O N a\right] = {n}_{\text{sodium acetate")/V_("total") = n_("sodium acetate}} / \left({V}_{1} + {V}_{2}\right)$

The number of moles of sodium acetate can be found by using the initial concentration of $\text{50.0 mM}$

n_("sodium acetate") = "50.0 mM" * V_2 = "0.05" * V_2

Likewise, the concentration of acetic acid can be written as

$\left[C {H}_{3} C O O H\right] = {n}_{\text{acetic acid}} / \left({V}_{1} + {V}_{2}\right)$

The number of moles of acetic acid will equal

n_("acetic acid") = "50.0 mM" * V_1 = 0.05 * V_1#

Plug these values into equation (2) and you'll get

$\frac{0.05 \cdot {V}_{2}}{{V}_{1} + {V}_{2}} \cdot \frac{{V}_{1} + {V}_{2}}{0.05 \cdot {V}_{1}} = 0.155$, or

${V}_{2} / {V}_{1} = 0.155 \implies {V}_{2} = 0.155 \cdot {V}_{1}$ $\to$ plug this into equation (1)

${V}_{1} + 0.155 \cdot {V}_{1} = 0.2 \implies {V}_{1} = \frac{0.2}{1.155} = 0.173$ , and

${V}_{2} = 0.155 \cdot {V}_{1} = 0.155 \cdot 0.173 = 0.027$

These values represent the volumes in liters for the sodium acetate and acetic acid solutions you need to mix to get this buffer.

Therefore, the volume of acetic acetic solution you need is

${V}_{1} = \text{0.173 L" = "173 mL}$

Feb 3, 2015

173.9 ml is required.

$H A r i g h t \le f t h a r p \infty n s {H}^{+} + {A}^{-}$

$p H = p {K}_{a} + \log \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$

Because $\left[{A}^{-}\right]$ and $\left[H A\right]$ are equal this simplifies to:

$p H = p {K}_{a} + \log \left[{V}_{{A}^{-}} / \left[{V}_{H A}\right]\right]$

Where $V$ refers to volume

$3.95 = 4.76 + \log \left[{V}_{{A}^{-}} / {V}_{H A}\right]$

$\log \left[{V}_{{A}^{-}} / {V}_{H A}\right] = 3.95 - 4.76 = - 0.81$

${V}_{{A}^{-}} / \left({V}_{H A}\right) = 0.15$

And

${V}_{{A}^{-}} + {V}_{H A} = 200$

2 eqns, 2 unknowns.

${V}_{A}^{-} = 0.15 {V}_{H A}$

$S o :$

$0.15 {V}_{H A} + {V}_{H A} = 200$

$1.15 {V}_{H A} = 200$

${V}_{H A} = \frac{200}{1.15} = 173.9 m l$