If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K.

$\left[\mathrm{KOH}_{a q}\right]=\frac{0.561}{\frac{1}{5}} \mathrm{~g} / L$

$=2.805 \mathrm{~g} / L$

$=2.805 \times \frac{1}{56.11} \mathrm{M}$

$=.05 \mathrm{M}$

$\mathrm{KOH}_{(a q)} \rightarrow \mathrm{K}_{(a q)}^{+}+\mathrm{OH}^{-}{ }_{(a q)}$

$\left[\mathrm{OH}^{-}\right]=.05 \mathrm{M}=\left[\mathrm{K}^{+}\right]$

$\left[\mathrm{H}^{+}\right]\left[\mathrm{H}^{-}\right]=K_{\mathrm{w}}$

$\left[\mathrm{H}^{+}\right] \frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}$

$=\frac{10^{-14}}{0.05}=2 \times 10^{-13} \mathrm{M}$

$\therefore \mathrm{pH}=12.70$