What is the pH of a solution made by mixing 25.00 mL of 0.100 M HCl with 40.00 mL of 0.100 M KOH? Assume that the volumes of the solutions are additive.
A)0)64
B)1)64
C)12.36
D)13.36

Question

Quizplus · Accepted Answer

First, we need to calculate the number of moles of HCl and KOH that were added to the mixture:

n(HCl) = (0.100 mol/L) x (25.00 mL / 1000 mL/L) = 0.00250 mol
n(KOH) = (0.100 mol/L) x (40.00 mL / 1000 mL/L) = 0.00400 mol

The equation for the reaction between HCl and KOH is:

HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

Since KOH is a strong base, it will completely dissociate in water to form K+ and OH- ions. Therefore, the OH- ion concentration in the solution will be:

[OH-] = n(OH-) / V(total)
where n(OH-) is the number of moles of OH- ions formed during the reaction, and V(total) is the total volume of the solution (in liters).

Since KOH and HCl react in a 1:1 stoichiometry, the number of moles of OH- ions formed is equal to the number of moles of HCl initially added:

n(OH-) = n(HCl) = 0.00250 mol

The total volume of the solution is:

V(total) = V(HCl) + V(KOH) = 25.00 mL + 40.00 mL = 65.00 mL = 0.065 L

Therefore,

[OH-] = n(OH-) / V(total) = 0.00250 mol / 0.065 L = 0.0385 mol/L

From this, we can calculate the pOH of the solution:

pOH = -log[OH-] = -log(0.0385) = 1.41

Finally, we can use the fact that pH + pOH = 14 to calculate the pH of the solution:

pH = 14 - pOH = 14 - 1.41 = 12.36

Therefore, the pH of the solution is 12.36. The answer is (C).

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