Question 17.13: Calculating the pH of a Solution of a Strong Acid and a Stro...
Calculating the pH of a Solution of a Strong Acid and a Strong Base
Calculate the \mathrm{pH} of a solution in which 10.0 \mathrm{~mL} of 0.100 \mathrm{M} \mathrm{NaOH} is added to 25.0 \mathrm{~mL} of 0.100 \mathrm{M} \mathrm{HCl}.
Because the reactants are a strong acid and a strong base, the reaction is essentially complete. The equation is
\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l)
You get the amounts of reactants by multiplying the volume (in liters) of each solution by its molar concentration.
\begin{aligned} & \text { Mol } \mathrm{H}_{3} \mathrm{O}^{+}=0.0250 \mathrm{~L} \times 0.100 \mathrm{~mol} / \mathrm{L}=0.00250 \mathrm{~mol} \\ & \mathrm{Mol} \mathrm{OH}^{-}=0.0100 \mathrm{~L} \times 0.100 \mathrm{~mol} / \mathrm{L}=0.00100 \mathrm{~mol} \end{aligned}
All of the \mathrm{OH}^{-}reacts, leaving an excess of \mathrm{H}_{3} \mathrm{O}^{+}.
\text { Excess } \mathrm{H}_{3} \mathrm{O}^{+}=(0.00250-0.00100) \mathrm{mol}=0.00150 \mathrm{~mol} \mathrm{H}_{3} \mathrm{O}^{+}
You obtain the concentration of \mathrm{H}_{3} \mathrm{O}^{+}by dividing this amount of \mathrm{H}_{3} \mathrm{O}^{+}by the total volume of solution (=0.0250 \mathrm{~L}+0.0100 \mathrm{~L}=0.0350 \mathrm{~L}).
\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{0.00150 \mathrm{~mol}}{0.0350 \mathrm{~L}}=0.0429
Hence,
\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log (0.0429)=\mathbf{1 . 3 6 8}