15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T6f4,

To determine the concentration of $$\mathrm{H}^{+}$$ ions in the resulting solution, we need to consider the contributions of $$\mathrm{HCl}$$ and $$\mathrm{H}_2\mathrm{SO}_4$$. Both provide $$\mathrm{H}^{+}$$ ions but in different proportions.

First, let’s break it down for each acid:

1. Hydrochloric Acid ($$\mathrm{HCl}$$):

$$\mathrm{HCl}$$ is a strong acid that dissociates completely in water:

$$\mathrm{HCl} \rightarrow \mathrm{H}^{+} + \mathrm{Cl}^{-}$$

The given concentration of $$\mathrm{HCl}$$ is $$0.1 M$$. Therefore, the concentration of $$\mathrm{H}^{+}$$ ions contributed by $$\mathrm{HCl}$$ in the solution is also $$0.1 M$$.

2. Sulfuric Acid ($$\mathrm{H_2SO_4}$$):

$$\mathrm{H_2SO_4}$$ is a strong acid and dissociates completely into two $$\mathrm{H}^{+}$$ ions and one $$\mathrm{SO_4}^{2-}$$ ion:

$$\mathrm{H_2SO_4} \rightarrow 2\mathrm{H}^{+} + \mathrm{SO_4}^{2-}$$

The given concentration of $$\mathrm{H_2SO_4}$$ is $$0.2 M$$. Each mole of $$\mathrm{H_2SO_4}$$ produces 2 moles of $$\mathrm{H}^{+}$$ ions. Therefore, the concentration of $$\mathrm{H}^{+}$$ ions contributed by $$\mathrm{H_2SO_4}$$ in the solution is:

$$0.2 M \times 2 = 0.4 M$$

When equal volumes of the two solutions are mixed, the concentration of $$\mathrm{H}^{+}$$ ions in the resulting solution is the average of the two contributions.

Hence, the overall concentration of $$\mathrm{H}^{+}$$ ions in the resulting solut

Ionic Equilibrum

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