# Evaluate mass of salt needed to add to a buffer solution knowing only pH

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.

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.