Compare the \(\mathrm{pH}\) of a \(0.040 \mathrm{M} \mathrm{HCl}\) solution with that of a \(0.040 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution. (Hint: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a strong acid; \(K_{\mathrm{a}}\) for \(\left.\mathrm{HSO}_{4}^{-}=1.3 \times 10^{-2} .\right)\)

Short Answer

Expert verified
The \(0.040 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution is more acidic and thus has a lower pH (approximately 1.1) compared to the\(0.040 \mathrm{M} \mathrm{HCl}\) solution whose pH is approximately 1.4.

Step by step solution

01

Understand the Dissociation of Strong Acids

Strong acids dissociate completely in solution. \(\mathrm{HCl}\) is a strong acid and it will donate one proton (\(\mathrm{H^{+}}\)) per molecule. Therefore, \(\mathrm{[H^{+}]}\) in the \(\mathrm{HCl}\) solution is 0.040 M. The \(\mathrm{H_{2}SO_{4}}\) is also a strong acid, but it can donate two protons (\(\mathrm{2H^{+}}\)) per molecule. That means, theoretically, in the absence of a second dissociation constant \(K_{a2}\), the \(\mathrm{[H^{+}]}\) of the \(\mathrm{H_{2}SO_{4}}\) solution would be 0.080 M which is twice as many protons as the \(\mathrm{HCl}\) solution.
02

Calculate the pH for Each Solution

The pH of a solution can be calculated using the formula \(pH = -log[H^{+}]\). For the \(\mathrm{HCl}\) solution, the calculation is \(pH = -log(0.040)\), which yields a pH of approximately 1.4. Likewise, for the \(\mathrm{H_{2}SO_{4}}\) solution, you will calculate the pH as \(pH = -log(0.080)\), which leads to a pH of approximately 1.1.
03

Compare the pH Levels

From the calculations in Step 2, you can see that the \(\mathrm{H_{2}SO_{4}}\) solution has a lower pH than the \(\mathrm{HCl}\) solution. Remember, lower pH values mean the solution is more acidic. This is consistent with the fact that \(\mathrm{H_{2}SO_{4}}\) donates more protons to the solution than \(\mathrm{HCl}\) does.

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