Question

What mass of KOH is necessary to prepare 800.0 mL of a solution having a pH = 11.56?

Short answer

0.163 g mass of KOH is required to prepare 800.0 mL of a solution having a pH = 11.56.

Step-by-step solution

OH- Concentration Calculation of KOH Solution

At 25°C,

$$\begin{gathered} \mathrm{pH}+\mathrm{pOH}=14 \\ \mathrm{pOH}=14-\mathrm{pH}=14-11.56=2.44 \\ \mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right] \\ \left[{\mathrm{OH}}^{-}\right]={10}^{-2.44}={10}^{-2.44}=3.63\times {10}^{-3} \end{gathered}$$

Calculation of Mass of KOH Solution

KOH is a strong base, and it dissociates completely.1 mol of KOH gives 1 moles of OH^- ions so,

$$\begin{gathered} \mathrm{Conc}.\mathrm{of}\mathrm{KOH}=\left[{\mathrm{OH}}^{-}\right]=3.63\times {10}^{-3}\mathrm{M} \\ \mathrm{Conc}.=\frac{\mathrm{Mole}}{\mathrm{volume}\left(\mathrm{L}\right)}=\frac{\mathrm{Mole}}{0.8}=3.63\times {10}^{-3} \end{gathered}$$

Mass = Mole × Molar Mass = × 56 = 0.163 g