Question 9.2: Calculate the pH of a 1.0 × 10^-3 M solution of HCl....
Calculate the \mathrm{pH} of a 1.0 \times 10^{-3} \mathrm{M} solution of \mathrm{HCl}.
Final Answer
\mathrm{HCl} is a strong acid. If 1 \mathrm{~mol} \,\mathrm{HCl} dissolves and dissociates in 1 \mathrm{~L} of aqueous solution, it produces 1 \mathrm{~mol} \,\mathrm{H}_{3} \mathrm{O}^{+}\left(\mathrm{a} 1 \mathrm{M}\right. solution of \left.\mathrm{H}_{3} \mathrm{O}^{+}\right). Therefore a 1.0 \times 10^{-3} \mathrm{M} \,\mathrm{HCl} solution has \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.0 \times 10^{-3} \mathrm{M}, and
\begin{aligned}\mathrm{pH} & =-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \\& =-\log \left[1.0 \times 10^{-3}\right]\end{aligned}
Consider the concentration term as composed of two parts, 1.0 and 10^{-3}. The logarithm of 1.0=0, and the logarithm of 10^{-3} is simply the exponent, -3 . Therefore
\begin{aligned}\mathrm{pH} & =-\left[\log 1.0+\log 10^{-3}\right] \\& =-[0-3.00] \\& =-[-3.00]=3.00\end{aligned}