We are asked to determine the pH of the solution.

First, let's calculate the moles of HCl and dibasic base B

$\mathbf{moles}\mathbf{}\mathbf{B}\mathbf{}\mathbf{=}\mathbf{}\mathbf{50}\mathbf{}\overline{)\mathbf{mL}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}\mathbf{\times}\mathbf{}\frac{\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{}\mathbf{mol}}{\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}$** = 0.0475 moles**

**$\mathbf{moles}\mathbf{}{\mathbf{H}}^{\mathbf{+}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{50}\mathbf{}\overline{)\mathbf{mL}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}\mathbf{\times}\mathbf{}\frac{\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{}\mathbf{mol}}{\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}$ = 0.0475 moles**

The moles of HCl and moles of B are equivalent. This means that we reached the equivalence point.

The pKb values for the dibasic base B are p_{Kb1} = 2.1 and p_{Kb2} = 7.5.

Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.95 M B(aq) with 0.95 M HCl(aq).

(c) after addition of 50.0 mL of HCl

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Equivalence Point concept. If you need more Equivalence Point practice, you can also practice Equivalence Point practice problems.