# What is the pH of a buffer solution prepared from 50*mL of sodium hydroxide solution that is 0.220*mol*L^-1 with respect to NaOH, and 100*mL of acetic acid solution that is 0.150*mol*L^-1 with respect to HOAc?

Mar 16, 2017

$p H = 5.20$

#### Explanation:

We use the buffer equation such that,

pH=pK_a+log_10{[[""^(-)OAc]]/[[HOAc]]}

But of course, we have to calculate [""^(-)OAc] and $\left[H O A c\right]$. The volume of the solution is clearly $150 \cdot m L$.

$\text{Moles of NaOH} \equiv \left(50.0 \times {10}^{-} 3 L \times 0.220 \cdot m o l \cdot {L}^{-} 1\right) = 0.011 \cdot m o l .$

$\text{Moles of HOAc} \equiv \left(100.0 \times {10}^{-} 3 L \times 0.150 \cdot m o l \cdot {L}^{-} 1\right) = 0.015 \cdot m o l .$

And since $N a O H$ reacts quantitatively with $H O A c$, we have a solution that is nominally,

$\left[H O A c\right] = \frac{\left(0.015 - 0.011\right) \cdot m o l}{150 \times {10}^{-} 3 L} = 0.0267 \cdot m o l \cdot {L}^{-} 1$

And [""^(-)OAc]=(0.011*mol)/(150xx10^-3L)=0.0733*mol*L^-1

And so we fill in the dots; the $p H$ should be elevated from the $p {K}_{a}$ (which I happen to know is $4.76$; these data really should have been supplied with the question!).

$p H = 4.76 + {\log}_{10} \left\{\frac{0.0733 \cdot m o l \cdot {L}^{-} 1}{0.0267 \cdot m o l \cdot {L}^{-} 1}\right\}$

$= 4.76 + 0.439 = 5.20$.