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

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

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