91Ó°ÊÓ

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

The equation to find \(pOH\)is\(pH + pOH = p{K_w}\)

Hence, the ionization of \(HCl\)is,

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

Since, it is given that, the concentration of \(HCl\)is \(2.0mol{L^{ - 1}}\), the concentration can be written as,

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

Thus,

\(c\left( {{H^ + }} \right) = 2.0mol{L^{ - 1}}\)

Calculate the \(pH\)by using the formula,

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

Substitute \(c\left( {{H^ + }} \right) = 2.0mol{L^{ - 1}}\)

\(\begin{aligned}{pH = - \log \frac{{2.0mol{L^{ - 1}}}}{{mol{L^{ - 1}}}}}\\{pH = - 0.3}\end{aligned}\)

Therefore, the value of\(pH\)is\( - 0.3\).

Use the equation,

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

Since,

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

Hence, the ionization of water is\({\rm{\;}}1.0 \cdot {10^{ - 14}}\) .

Substitute, \({{\rm{H}}^ + }{\rm{ = \;}}1.0 \cdot {10^{ - 14}}\) .

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

Calculate\(pOH\),

\(\begin{aligned}{pOH = p{K_w} - pH}\\{pOH = 14 + 0.3}\\{pOH = 14.3}\end{aligned}\)

Therefore, the value of\(pOH\)is\(14.3\).