Calculate \(pH\;and the\;pOH\) of each of the following solutions at\(2{5^o}C\)for which the substances ionize completely:

(a)\(0.200M HCl\)

(b)\(0.0143M NaOH\)

(c)\(3.0M HN{O_3}\)

(d) \(0.0031M Ca{(OH)_2}\)

The formula to find\(pH\)is,\(pH = - \log \left( {\frac{{c\left( {{H^ + }} \right)}}{{mol{L^{ - 1}}}}} \right)\)

The formula to find\(pOH\)is,\(pOH = - \log \left( {\frac{{c\left( {O{H^ - }} \right)}}{{mol{L^{ - 1}}}}} \right)\)

The equation\(pH + pOH = p{K_w}\)can be used to find both\(pH\)and\(pOH.\)

Consider the reaction of ionization of \(HCl,\)

\(HCl(aq) \to {H^ + }(aq) + C{l^ - }(aq)\)

Since, it is given, the concentration of \(HCl\)is\(0.2{\rm{mol}}{L^{ - 1}}\), it can be written as,

\(c(HCl) = c\left( {{H^ + }} \right)\)

Thus,

\(c\left( {{H^ + }} \right) = 0.2\frac{{mol}}{L}\)

Calculate the\(pH\)of the solution,

\(pH = - \log \left( {\frac{{c\left( {{H^ + }} \right)}}{{mol{L^{ - 1}}}}} \right)\)

Substitute, \(c\left( {{H^ + }} \right) = 0.2\frac{{mol}}{L}\)

\(\begin{aligned}{pH = - \log \left( {\frac{{0.2{\rm{mol}}{{\rm{L}}^{ - 1}}}}{{{\rm{mol}}{{\rm{L}}^{ - 1}}}}} \right)}\\{}\end{aligned}\)

\(pH = 0.7\)

Use the equation, to find the\(pOH\),

\(pH + pOH = p{K_w}\)

Since,\(p{K_w} = - \log {K_w}\)

Substitute, the ionization constant of water\(({K_w}) = 1.0 \times {10^{ - 14}}\)

\(\begin{aligned}{p{K_w} = - \log \left( {1.0 \cdot {{10}^{ - 14}}} \right)}\\{}\end{aligned}\)

\(p{K_w} = 14\)

Thus,

\(pH + pOH = p{K_w}\)

\(\begin{aligned}{pOH = 14 - 0.7}\\{pOH = 13.3}\end{aligned}\)

Therefore, \({\rm{pH\;and the\;pOH}}\)of the solution \(HCl\) at \({25^ \circ }C\)is \(pH = 0.7\)and\(pOH = 13.3\)

Consider the reaction of ionization of \(NaOH,\)

\(NaOH(aq) \to N{a^ + }(aq) + O{H^ - }(aq)\)

Since, the reaction is complete, the equation can be written as,

\(c(NaOH) = c\left( {O{H^ - }} \right)\)

Hence, it is given that the concentration of\(NaOH\)is\(0.0143\frac{{mol}}{L}\),

Thus,\(c\left( {O{H^ - }} \right) = 0.0143\frac{{{\rm{mol}}}}{L}\)

Calculate the\(pOH\),

\(pOH = - \log \left( {\frac{{c\left( {O{H^ - }} \right)}}{{mol{L^{ - 1}}}}} \right)\)

\(\begin{aligned}{pOH = - \log \left( {\frac{{0.0143{\rm{mol}}{{\rm{L}}^{ - 1}}}}{{mol{L^{ - 1}}}}} \right)}\\{pOH = 1.84}\end{aligned}\)

Solve for\(pH\),

\(\begin{aligned}{pH + pOH = p{K_w}}\\{pH = 14 - 1.84}\\{pH = 12.16}\end{aligned}\)

Therefore, \({\rm{pH\;and the\;pOH}}\)of the solution \(NaOH\)at \({25^ \circ }C\)is \(pH = 12.16\)and the \(pOH = \)\(1.84\).

Consider the reaction of ionization of\(HN{O_3}\),

\({\rm{HN}}{{\rm{O}}_3}(aq) \to {{\rm{H}}^ + }(aq) + {\rm{NO}}_3^ - (aq)\)

Since, reaction is complete, it can be written as,

\(c\left( {{\rm{HN}}{{\rm{O}}_3}} \right) = c\left( {{{\rm{H}}^ + }} \right)\)

Hence, it is given that, the concentration of\(HN{O_3}\)is\(3.0mol{L^{ - 1}}\)

\(c\left( {{H^ + }} \right) = 3.0\frac{{{\rm{mol}}}}{L}\)

Calculate\(pH\)of the solution,

\(pH = - \log \left( {\frac{{c\left( {{H^ + }} \right)}}{{mol{L^{ - 1}}}}} \right)\)

\(\begin{aligned}{pH = - \log \left( {\frac{{3.0mol{L^{ - 1}}}}{{mol{L^{ - 1}}}}} \right)}\\{pH = - 0.5}\end{aligned}\)

Use the equation, to find the\(pOH\),

\(pH + pOH = p{K_w}\)

Since,\(p{K_w} = 14\), the value of\(pOH\)can be calculated as,

\(\begin{aligned}{pOH = 14 + 0.5}\\{pOH = 14.5}\end{aligned}\)

Therefore, \({\rm{pH\;and the\;pOH}}\)of the solution \(HN{O_3}\) at \({25^ \circ }C\) is \(pH = - 0.5\)and the \(pOH = \)\(14.5\).

Consider the reaction of ionization of\(Ca{(OH)_2}\),

\(Ca{(OH)_2}(aq) \to C{a^{2 + }}(aq) + 2O{H^ - }(aq)\)

Since, the reaction is complete, it can be written as,

\(c\left( {Ca{{(OH)}_2}} \right):c\left( {O{H^ - }} \right) = 1:2\)

\(2c\left( {Ca{{(OH)}_2}} \right) = c\left( {O{H^ - }} \right)\)

Hence, it is given that, the concentration of\(Ca{(OH)_2}\)is\(0.0031\frac{{{\rm{mol}}}}{L}\)

\(\begin{aligned}{c\left( {O{H^ - }} \right) = 2 \cdot 0.0031\frac{{mol}}{L}}\\{c\left( {O{H^ - }} \right) = 0.0062\frac{{mol}}{L}}\end{aligned}\)

Calculate\(pOH\)of the solution,

\(pOH = - \log \left( {\frac{{c\left( {O{H^ - }} \right)}}{{mol{L^{ - 1}}}}} \right)\)

\(\begin{aligned}{pOH = - \log \left( {\frac{{0.0062mol{L^{ - 1}}}}{{mol{L^{ - 1}}}}} \right)}\\{pOH = 2.2}\end{aligned}\)

Use the equation to find\(pH\),

\(\begin{aligned}{pH + pOH = p{K_w}}\\{pH = 14 - 2.2}\\{pH = 11.8}\end{aligned}\)

Therefore, \({\rm{pH\;and the\;pOH}}\)of the solution\(Ca{(OH)_2}\)at\({25^ \circ }C\)is\(pH = 11.8\)and the\(pOH = \)\(2.2\)