(A) The formula for $pH$ is defined as $pH = - \log [H^{+}]$.Given that $[H^{+}] = 0.01 \, M$,which can be written as $10^{-2} \, M$.Substituting the

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(A) The formula for $pH$ is defined as $pH = - \log [H^{+}]$.Given that $[H^{+}] = 0.01 \, M$,which can be written as $10^{-2} \, M$.Substituting the value into the formula: $pH = - \log [10^{-2}]$.Using the logarithmic property $\log(a^b) = b \log(a)$,we get $pH = -(-2) \log(10)$.Since $\log(10) = 1$,the $pH = 2$.