What is the pH of a solution made by mixing 25.00 mL of 0. 100 mol L^-1 HCl with 40.00 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

Question

Quizplus · Accepted Answer

Step 1: Determine the initial moles of acid and base

moles HCl = (0.100 mol/L) * (25.00 mL/1000 mL) = 0.00250 mol 
moles KOH = (0.100 mol/L) * (40.00 mL/1000 mL) = 0.00400 mol

Step 2: Determine which reagent is in excess

From the balanced chemical equation for the reaction between HCl and KOH:

HCl + KOH → H2O + KCl

It can be seen that one mole of HCl reacts with one mole of KOH to produce one mole of water and one mole of KCl. Therefore, the limiting reagent will be the one that has fewer moles.

Since HCl has 0.00250 mol and KOH has 0.00400 mol, HCl is the limiting reagent.

Step 3: Determine the moles of remaining reagent after reaction

The reaction consumes all of the HCl, leaving 0 moles of HCl.

Step 4: Determine the concentration of the remaining reagent

Since there are 0 moles of HCl remaining and 0.00400 moles of KOH initially, there are still 0.00150 moles of KOH remaining.

The final volume of the solution is the sum of the initial volumes:

Vfinal = 25.00 mL + 40.00 mL = 65.00 mL

The final concentration of KOH can be calculated as follows:

[OH-] = moles KOH / Vfinal

[OH-] = 0.00150 mol / 0.06500 L = 0.0231 M

The pOH of the solution is:

pOH = -log[OH-]

pOH = -log(0.0231) = 1.636

The pH of the solution can be calculated as:

pH = 14 - pOH

pH = 14 - 1.636 = 12.36

Therefore, the pH of the solution is 12.36, which corresponds to choice C.

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