1
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.