Calculating the pH of a Solution of a Strong Acid and a Strong Base Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl.

Question

Calculating the pH of a Solution of a Strong Acid and a Strong Base

Calculate the [latex]\mathrm{pH}[/latex] of a solution in which [latex]10.0 \mathrm{~mL}[/latex] of [latex]0.100 \mathrm{M} \mathrm{NaOH}[/latex] is added to [latex]25.0 \mathrm{~mL}[/latex] of [latex]0.100 \mathrm{M} \mathrm{HCl}[/latex].

Accepted Answer

Because the reactants are a strong acid and a strong base, the reaction is essentially complete. The equation is

[latex]\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l)[/latex]

You get the amounts of reactants by multiplying the volume (in liters) of each solution by its molar concentration.

[latex] \begin{aligned} & ext { Mol } \mathrm{H}_{3} \mathrm{O}^{+}=0.0250 \mathrm{~L} imes 0.100 \mathrm{~mol} / \mathrm{L}=0.00250 \mathrm{~mol} \ & \mathrm{Mol} \mathrm{OH}^{-}=0.0100 \mathrm{~L} imes 0.100 \mathrm{~mol} / \mathrm{L}=0.00100 \mathrm{~mol} \end{aligned} [/latex]

All of the [latex]\mathrm{OH}^{-}[/latex]reacts, leaving an excess of [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex].

[latex] ext { Excess } \mathrm{H}_{3} \mathrm{O}^{+}=(0.00250-0.00100) \mathrm{mol}=0.00150 \mathrm{~mol} \mathrm{H}_{3} \mathrm{O}^{+}[/latex]

You obtain the concentration of [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex]by dividing this amount of [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex]by the total volume of solution [latex](=0.0250 \mathrm{~L}+0.0100 \mathrm{~L}=0.0350 \mathrm{~L})[/latex].

[latex]\left[\mathrm{H}_{3} \mathrm{O}^{+} ight]=\frac{0.00150 \mathrm{~mol}}{0.0350 \mathrm{~L}}=0.0429[/latex]

Hence,

[latex]\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+} ight]=-\log (0.0429)=\mathbf{1 . 3 6 8}[/latex]

Question 17.13: Calculating the pH of a Solution of a Strong Acid and a Stro...

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