What is the pH of a solution made by mixing 40.00 mL of 0.100 mol L^-1 HCl with 25.00 mL of 0.100 mol L^-1 KOH? Assume that the volumes of the solutions are additive. A) 0.64 B) 1.64 C) 12.36 D) 13.36 E) 10.00

Question

Quizplus · Accepted Answer

First, we need to find the amount of moles of HCl and KOH in the solution:

moles HCl = (0.100 mol/L) x (0.04000 L) = 0.00400 mol
moles KOH = (0.100 mol/L) x (0.02500 L) = 0.00250 mol

Since KOH is a strong base, it will completely dissociate in solution according to the following reaction:

KOH + H2O → K+ + OH-

This means that the concentration of OH- in the solution will be 0.00250 mol / 0.02500 L = 0.100 mol/L

The reaction between HCl and KOH will result in the net reaction:

HCl + KOH → KCl + H2O

This means that the amount of H+ in the solution will be equal to the amount of OH-, since the reaction produces 1 mol of water for every 1 mol of HCl and KOH reacted. Therefore, the concentration of H+ in the solution will also be 0.100 mol/L.

We can use the formula for pH to calculate the pH of the solution:

pH = -log[H+]

pH = -log(0.100) = 1.00

However, we need to remember that we mixed 40.00 mL of HCl with 25.00 mL of KOH. Since the volumes are additive, the total volume of the solution is:

Vtotal = 0.04000 L + 0.02500 L = 0.06500 L

To calculate the final pH of the solution, we need to recalculate the concentration of H+ based on the total volume:

[H+] = (moles H+) / Vtotal

moles H+ = moles OH- = 0.00250 mol

[H+] = 0.00250 mol / 0.06500 L = 0.0385 mol/L

pH = -log(0.0385) = 1.41

Therefore, the pH of the solution is 1.41, which corresponds to answer choice B.

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