**$\mathbf{550}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{mL}}\mathbf{}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}\mathbf{\times}\frac{\mathbf{0}\mathbf{.}\mathbf{120}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{HCl}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{H}}^{\mathbf{+}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{HCl}}}$ = 0.066 mol H ^{+}**

**$\mathbf{550}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{mL}}\mathbf{}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}\mathbf{\times}\frac{\mathbf{0}\mathbf{.}\mathbf{210}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{H}}^{\mathbf{+}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}}$ = 0.231 mol H ^{+}**

An acid solution is 0.120 M in HCl and 0.210 M in H_{2} SO_{4}.

What volume of a 0.160 M solution of KOH must be added to 550.0 mL of the acidic solution to completely neutralize all of the acid?

Frequently Asked Questions

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 Solution Stoichiometry concept. You can view video lessons to learn Solution Stoichiometry. Or if you need more Solution Stoichiometry practice, you can also practice Solution Stoichiometry practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Brydges' class at UCSD.