0
$\begingroup$

This is an exercise taken from an old exam, I'm struggling with the resolution.

To a solution of a generic weak acid $\ce{HA}$ were added $\pu{2.40 g}$ of a potassium salt of the $\ce{KA}$. The solution has $\mathrm{pH} = 4.8$. Evaluate how many grams of salt are needed if we want to shift the solution to a $\mathrm{pH} = 5$.

Neither $K_\mathrm{a}$ nor $K_\mathrm{b}$ are given, nor anything. The only thing I got is just confusion.

$\endgroup$
1
$\begingroup$

I think there is all the data you need. $\mathrm{pH}$ of a buffer formed by a weak acid $\ce{HA}$ and its potassium salt $\ce{KA}$ can be found as

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{C(\ce{KA})}{C(\ce{HA})}}$$

On the other hand

$$C(\ce{KA}) = \frac{m(\ce{KA})}{M(\ce{KA})V}$$

where $m$ and $M$ are mass and molecular mass of $\ce{KA}$; $V$ is volume. In general, first equation can be rewritten as

$$ \begin{align} \mathrm{pH}_i &= \mathrm{p}K_\mathrm{a} + \log{C_i(\ce{KA})} - \log{C(\ce{HA})} \\ &= \mathrm{p}K_\mathrm{a} + \log{m_i(\ce{KA})} -\log{M(\ce{KA})} - \log{V} - \log{C(\ce{HA})} \end{align} $$

Rearranging:

$$\mathrm{pH}_i - \log{m_i(\ce{KA})} = \mathrm{p}K_\mathrm{a} -\log{M(\ce{KA})} - \log{V} - \log{C(\ce{HA})} = \mathrm{const}$$

so that now we can equate conditions for both solutions and find the mass:

$$\mathrm{pH}_1 - \log{m_1(\ce{KA})} = \mathrm{pH}_2 - \log{m_2(\ce{KA})}$$ $$\log{\frac{m_2(\ce{KA})}{m_1(\ce{KA})}} = \mathrm{pH}_2 - \mathrm{pH}_1$$ $$m_2(\ce{KA}) = m_1(\ce{KA})\cdot10^{\mathrm{pH}_2 - \mathrm{pH}_1} = \pu{2.40 g}\cdot10^{5.0-4.8}\approx\pu{3.80 g}$$

So, in order to achieve $\mathrm{pH} = 5.0$, one has to add $\pu{3.80 g} - \pu{2.40 g} = \pu{1.40 g}$ of potassium salt.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

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