# Computing pH of a buffer, and concentrations

You pour $50~\mathrm{mL}$ of a $0.0200~\mathrm{M}$ $\ce{HCOOH}$-solution to $150~\mathrm{mL}$ of a $0.0500~\mathrm{M}$ $\ce{HCOONa}$-solution. The $K_\mathrm{a}$ of $\ce{HCOOH}$ is $1.8 \times 10^{-4}$ and $\mathrm{p}K_\mathrm{a}$ is $3.74$.

a) Compute the concentrations of $\ce{HCOOH}$ and its salt after dilution to this $200~\mathrm{mL}$ buffer solution.

b) Calculate the $\mathrm{pH}$ of this buffer solution with the buffer formula.

Attempt at solution: Not sure how to do part a). For part b) I used the following method. We have $50 \times 10^{-3}~\mathrm{L} \cdot \frac{0.0200~\mathrm{mol}}{1~\mathrm{L}} = 0.001~\mathrm{mol}$ of $\ce{HCOOH}$. Doing this again for $\ce{HCOONa}$ gives $0.0075~\mathrm{mol}$. Hence: \begin{align*} \mathrm{pH} = 3.74 + \log(\frac{0.0075}{0.001}) = 4.61 \end{align*}

But how can I solve a)? I'm having trouble with writing the reaction down. I'm not sure what's happening. Which chemical is neutralizing which? And what are the ions?

We have to compute the new concentrations after dilution: $$\ce{[HCOOH]}=\frac{0.02\times 50}{200}=0.005\,\, \mathrm{mol/L}$$ $$\ce{[HCOO^- ]}=\frac{0.05\times 150}{200}=0.0375\,\, \mathrm{mol/L}$$
Now, we can compute the $\mathrm{pH}$ of the buffer solution:$$\mathrm{pH}=\mathrm{p}K_a +\log \frac{\ce{[HCOO^- ]}}{\ce{[HCOOH]}}$$
$$\mathrm{pH}=3.74+\log \frac{0.0375}{0.005}=4.62$$
• Thanks for the help. How would I write this reaction down? $HCOOH \rightarrow HCOO^- + H^+$ ? And what does the 'Na' in front of the HCOO mean? – Kamil Apr 28 '15 at 13:58