# Question #35ef6

Mar 25, 2017

You start from the buffer equation:

$p H = p {K}_{a} + {\log}_{10} \left(\frac{\left[{C}_{3} {H}_{7} {O}_{2}^{-}\right]}{\left[{C}_{3} {H}_{7} C {O}_{2} H\right]}\right)$

#### Explanation:

And for your problem:

$5.17 = 4.069 + {\log}_{10} \left(\frac{\left[{C}_{3} {H}_{7} {O}_{2}^{-}\right]}{\left[{C}_{3} {H}_{7} C {O}_{2} H\right]}\right)$

Clearly, ${\log}_{10} \left(\frac{\left[{C}_{3} {H}_{7} {O}_{2}^{-}\right]}{\left[{C}_{3} {H}_{7} C {O}_{2} H\right]}\right) = 5.17 - 4.069 = 1.10$

And we take antilogs.........

${10}^{1.10} = \frac{\left[{C}_{3} {H}_{7} {O}_{2}^{-}\right]}{\left[{C}_{3} {H}_{7} C {O}_{2} H\right]} = 12.62$

But by spec. $\left[{C}_{3} {H}_{7} C {O}_{2} H\right] = 0.50 \cdot m o l \cdot {L}^{-} 1$.

So $\left[{C}_{3} {H}_{7} {O}_{2}^{-}\right] = 6.31 \cdot m o l \cdot {L}^{-} 1$

For convenience, we would first prepare a $6.81 \cdot m o l \cdot {L}^{-} 1$ solution of $\text{propionic acid}$ (which as I recall would not have a very pleasant smell!), and then add a $6.31 \cdot m o l$ $K O H$ to a $1 \cdot L$ volume.

Alternatively, we could add $6.31 \cdot m o l \times 96.07 \cdot g \cdot m o {l}^{-} 1 = 606.1 \cdot g$ $\text{sodium propionate}$ to a $1 \cdot L$ volume of $0.500 \cdot m o l \cdot {L}^{-} 1$ $\text{propionic acid}$. (The volumes would not change too much!)