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

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

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.