What are the hydronium and hydroxide ion concentrations in a solution whose \(pH\) is \(6.52?\)

The concentration of hydronium ions can be found from the\(pH\)with the equation,\(c\left( {{H_3}{O^ + }} \right) = {10^{ - pH}}mol{L^{ - 1}}\)

The concentration of hydroxide ions can be found from the \(pOH\) with the equation\(c\left( {O{H^ - }} \right) = {10^{ - pOH}}mol{L^{ - 1}}\).

Hence, the equation to find the concentration of hydronium ions is, \(c\left( {{H_3}{O^ + }} \right) = {10^{ - pH}}mol{L^{ - 1}}\)

Substitute the given \(pH\)value,

\(pH = 6.52\)

Thus,

\(c\left( {{H_3}{O^ + }} \right) = {10^{ - pH}}mol{L^{ - 1}}\)

\(\begin{aligned}{c\left( {{{\rm{H}}_3}{{\rm{O}}^ + }} \right) = {{10}^{ - 6.52}}{\rm{mol}}{{\rm{L}}^{ - 1}}}\\{c\left( {{{\rm{H}}_3}{{\rm{O}}^ + }} \right) = 3.02 \cdot {{10}^{ - 7}}{\rm{mol}}{{\rm{L}}^{ - 1}}}\end{aligned}\)

Since, the equation to find the concentration of hydroxide ion is, \(c\left( {O{H^ - }} \right) = {10^{ - pOH}}mol{L^{ - 1}}\).

Calculate the\(pOH\)using the equation,

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

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

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

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

Solve for\(pOH\),

\(\begin{aligned}{pOH = p{K_w} - pH}\\{pOH = 14 - 6.52}\\{pOH = 7.48}\end{aligned}\)

Substitute\(pOH = 7.48\)in the equation, \(c\left( {O{H^ - }} \right) = {10^{ - pOH}}mol{L^{ - 1}}\).

\(\begin{aligned}{c\left( {O{H^ - }} \right) = {{10}^{ - 7.48}}{\rm{mol}}{{\rm{L}}^{ - 1}}}\\{c\left( {O{H^ - }} \right) = 3.31 \cdot {{10}^{ - 8}}{\rm{mol}}{{\rm{L}}^{ - 1}}}\end{aligned}\)

Therefore, the concentration of hydronium ion is \(c\left( {{{\rm{H}}_3}{{\rm{O}}^ + }} \right) = 3.02 \cdot {10^{ - 7}}{\rm{mol}}{{\rm{L}}^{ - 1}}\)and the concentration of hydroxide ion is\(c\left( {O{H^ - }} \right) = 3.31 \cdot {10^{ - 8}}{\rm{mol}}{{\rm{L}}^{ - 1}}\).