Question

A 0.25-g sample of lime (CaO) is dissolved in enough water to make 1500 mL of solution. Calculate the pH of the solution.

Short answer

The pH of 0.25 g sample of lime (CaO) is dissolved in enough water to make 1500 mL of solution = 9.8.

Step-by-step solution

Amount of Calcium Hydroxide Prepared

$\begin{array}{l}\text{Moles\hspace{0.17em}of\hspace{0.17em}CaO}=\frac{0.25}{\text{56}}=4.46\times {10}^{-3}\\ \text{Conc\hspace{0.17em}in\hspace{0.17em}mole/L=}\frac{4.46\times {10}^{-3}}{\text{1.5}}\\ \text{Conc\hspace{0.17em}in\hspace{0.17em}mole/L}=2.97\times {10}^{-3}\end{array}$

CaO when get dissolved in water produces Ca(OH)₂.

data-custom-editor="chemistry" ${\text{CaO+H}}_{\text{2}}\text{O}\rightleftarrows {{\text{\hspace{0.17em}Ca(OH)}}_{\text{2}}}_{\text{(aq)}}$

From the reaction, we can see that 1 mole of CaO will give 1 mole of Ca(OH)₂.

Moles of CaO =Moles of Ca(OH)₂

1 mole of Ca(OH)₂ gives 2 moles of OH^-.

$\begin{array}{l}\text{Moles\hspace{0.17em}of\hspace{0.17em}}[{\text{OH}}^{\text{-}}\text{]}=2\times 2.97\times {10}^{-3}\\ \text{Moles\hspace{0.17em}of\hspace{0.17em}}[{\text{OH}}^{\text{-}}\text{]}=5.94\times {10}^{-3}\end{array}$

Calculation of pH

First of all lets calculate Ka

$\begin{array}{l}{\text{H}}^{\text{+}}=\frac{{\text{K}}_{\text{w}}}{\left[{\text{OH}}^{\text{-}}\right]}\\ {\text{H}}^{\text{+}}=\frac{1\times {10}^{-14}}{5.94\times {10}^{-3}}\\ {\text{H}}^{\text{+}}=1.68\times {10}^{-10}\end{array}$

$\begin{array}{l}{\text{pH=-log[H}}^{\text{+}}\text{]}\\ \text{pH=-log(1}.68\times {10}^{-10}\text{)}\\ \text{pH=9.8}\end{array}$