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

(a)\(0.000259M HCl{O_4}\)

(b)\(0.21M NaOH\)

(c)\(0.000071M Ba{(OH)_2}\)

(d) \(2.5M KOH\)

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 \(HCl{O_4}\),

\({\rm{HCl}}{{\rm{O}}_4}(aq) \to {{\rm{H}}^ + }(aq) + {\rm{ClO}}_4^ - (aq)\)

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

\(c\left( {HCl{O_4}} \right) = c\left( {{H^ + }} \right)\)

Since, it is given that, the concentration of \(HCl{O_4}\)is\(0.000259\,mol{L^{ - 1}}\)

\(c\left( {{H^ + }} \right) = 0.000259\frac{{mol}}{L}\)

Calculate\(pH\)of a solution,

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

Substitute \(c\left( {{H^ + }} \right) = 0.000259\frac{{mol}}{L}\)

\(\begin{aligned}{pH = - \log \left( {\frac{{0.000259mol{L^{ - 1}}}}{{mol{L^{ - 1}}}}} \right)}\\{pH = 3.6}\end{aligned}\)

Use the equation to find\(pOH,\)

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

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

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

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

Thus,

\(\begin{aligned}{pOH = 14 - 3.6}\\{pOH = 10.4}\end{aligned}\)

Therefore, value of \(pH\)and \(pOH\)of a solution \(HCl{O_4}\)is\(pH = 3.6\)and\(pOH = 10.4\).

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

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

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

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

Hence, the concentration of\(NaOH\)is given as\(c(NaOH) = 0.21\frac{{mol}}{L}\)

Thus,

\(c\left( {O{H^ - }} \right) = 0.21\frac{{mol}}{L}\)

Calculate the value of\(pOH\),

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

Substitute \(c\left( {O{H^ - }} \right) = 0.21\frac{{mol}}{L}\)

\(\begin{aligned}{pOH = - \log \left( {\frac{{0.21{\rm{mol}}{{\rm{L}}^{ - 1}}}}{{mol{L^{ - 1}}}}} \right)}\\{pOH = 0.68}\end{aligned}\)

Find the value of\(pH\),

\(\begin{aligned}{pH + pOH = p{K_w}}\\{pH = 14 - 0.68}\\{pH = 13.32}\end{aligned}\)

Therefore, the value of \(pH\)and \(pOH\)of a solution\(NaOH\)is\(pH = 13.32\)and\(pOH = 0.68\).

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

\({\rm{Ba}}{({\rm{OH}})_2}(aq) \to {\rm{B}}{{\rm{a}}^{2 + }}(aq) + 2{\rm{O}}{{\rm{H}}^ - }(aq)\)

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

\(\begin{aligned}{c\left( {Ba{{(OH)}_2}} \right):c\left( {O{H^ - }} \right) = 1:2}\\{2c\left( {Ba{{(OH)}_2}} \right) = c\left( {O{H^ - }} \right)}\end{aligned}\)

Hence, the concentration of\(Ba{(OH)_2}\)is given as\(0.000071\frac{{{\rm{mol}}}}{L}\).

Thus,

\(\begin{aligned}{c\left( {O{H^ - }} \right) = 2 \cdot 0.000071\frac{{{\rm{mol}}}}{{\rm{L}}}}\\{c\left( {O{H^ - }} \right) = 0.000142\frac{{{\rm{mol}}}}{{\rm{L}}}}\end{aligned}\)

Calculate\(pOH\)of a solution,

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

Substitute \(c\left( {O{H^ - }} \right) = 0.000142\frac{{{\rm{mol}}}}{{\rm{L}}}\)

\(\begin{aligned}{pOH = - \log \left( {\frac{{0.000142mol{L^{ - 1}}}}{{mol{L^{ - 1}}}}} \right)}\\{pOH = 3.85}\end{aligned}\)

Find \(pH\)of a solution,

\(\begin{aligned}{pH + pOH = p{K_w}}\\{pH = 14 - 3.85}\\{pH = 10.15}\end{aligned}\)

Therefore, the value of \(pH\)and the\(pOH\)of a solution \(Ba{(OH)_2}\)is\(pH = 10.15\)and \(pOH = 3.85\).

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

\(KOH(aq) \to {K^ + }(aq) + O{H^ - }(aq)\)

Since, the reaction is complete, we can write as

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

Hence, the concentration of\(KOH\)is given as\(2.5\frac{{{\rm{mol}}}}{{\rm{L}}}\).

Thus,

\(c\left( {O{H^ - }} \right) = 2.5\frac{{mol}}{L}\)

Calculate the value of\(pOH\),

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

Substitute \(c\left( {O{H^ - }} \right) = 2.5\frac{{mol}}{L}\)

\(\begin{aligned}{pOH = - \log \left( {\frac{{2.5{\rm{mol}}{{\rm{L}}^{ - 1}}}}{{mol{L^{ - 1}}}}} \right)}\\{pOH = - 0.4}\end{aligned}\)

Find the value of\(pH\)of the solution,

\(\begin{aligned}{pH + pOH = p{K_w}}\\{pH = 14 + 0.4}\\{pH = 14.4}\end{aligned}\)

Therefore, the value of \(pH\)and \(pOH\)of a solution\(KOH\)is\(pH = 14.4\)and\(pOH = - 0.4\).