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?


2 Answers 2


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:$$\mathrm{pH}=\mathrm{p}K_a +\log \frac{\ce{[HCOO^- ]}}{\ce{[HCOOH]}}$$

$$\mathrm{pH}=3.74+\log \frac{0.0375}{0.005}=4.62$$

  • $\begingroup$ 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? $\endgroup$
    – Kamil
    Apr 28, 2015 at 13:58
  • $\begingroup$ sodium ion is the counter ion of HCOO- to have a neutral compound $\endgroup$ Apr 28, 2015 at 14:19

In a buffer solution, reactions primarily occur between the externally introduced species and the buffer constituents, HCOONa buffers the incoming H(+) proton from a strong acid and HCOOH buffers OH(-) from a base.

there is, as i believe, no other reactions involved here


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