(A) The concentration of hydroxide ions is $[OH^-] = 0.05 \ mol \ L^{-1} = 5 \times 10^{-2} \ mol \ L^{-1}$.$pOH = - \log[OH^-] = - \log(5 \times 10^{

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(A) The concentration of hydroxide ions is $[OH^-] = 0.05 \ mol \ L^{-1} = 5 \times 10^{-2} \ mol \ L^{-1}$.$pOH = - \log[OH^-] = - \log(5 \times 10^{-2}) = 2 - \log 5 = 2 - 0.699 = 1.301$.Since $pH + pOH = 14$ at $25^{\circ}C$,we have $pH = 14 - 1.301 = 12.699$.Since the $pH > 7$,the solution is basic.