Question

Equal volumes of three acid solutions of pH 3,4 and 5 are mixed in a vessel. What will be the ${{\text{H}}^{\text{+}}}$ ion concentration in the mixture?
(A) 1.11×${{10}^{-4}}$M
(B) 3.7×${{10}^{-4}}$M
(C) 3.7×${{10}^{-3}}$ M
(D) 1.11×${{10}^{-3}}$ M

Answer 1

acidity of a solution iFor a solution containing different acids, the acidity of the solution is given by the average concentration of $\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}_{3}}{{\text{O}}^{\text{+}}}]$ in the solution.

Complete step by step solution:
Molarity of a solution is given by total number of moles per litre of solution or total number of millimoles in one mL solution
Therefore the final concentration is given by,

& \left[ {{\text{H}}^{\text{+}}} \right]\text{ from first acid }\\!\\!\times\\!\\!\text{ its volume + }\left[ {{\text{H}}^{\text{+}}} \right]\text{ from second acid }\\!\\!\times\\!\\!\text{ its volume + } \\\ & \left[ {{\text{H}}^{\text{+}}} \right]\text{from third acid }\\!\\!\times\\!\\!\text{ its volume} \\\ \end{aligned}}{\text{Total volume of the final solution}}$$ Therefore, Concentration in molarity of the final solution is given by, $$\begin{aligned} & \text{Molarity =}\dfrac{\text{mili Moles of solution}}{\text{Total volume in mL of the final solution}} \\\ & \text{ =}\dfrac{\text{Moles of solution}}{\text{Total volume in Litre of the final solution}} \\\ \end{aligned}$$ Assume the volume of each acidic solution is 1L therefore $$\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}_{3}}{{\text{O}}^{\text{+}}}]$$obtained from the solution of pH=3 is $${{10}^{-3}}$$ M $$\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}_{3}}{{\text{O}}^{\text{+}}}]$$obtained from from the solution of pH= 4 is $${{10}^{-4}}$$ M $$\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}_{3}}{{\text{O}}^{\text{+}}}]$$obtained from the solution of pH=5 is $${{10}^{-5}}$$M Thus Total concentration of $$\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}_{3}}{{\text{O}}^{\text{+}}}]$$ is given by , $$\text{Total }\\!\\![\\!\\!\text{ }{{\text{H}}_{3}}{{\text{O}}^{\text{+}}}]\text{ =}\dfrac{{{10}^{-3}}+{{10}^{-4}}+{{10}^{-5}}}{\text{ 3}}$$= $0.00037$=$3.7×$ $${{10}^{-4}}$$M. **Hence, answer to this question is option (C).** **Note:** Here the acids are not mentioned properly therefore we considered the acids to be monobasic. If they had mentioned then the solution could have given different answers. And if any of them were weak acid then their dissociation could have been suppressed.