Question

A 50.0 mL solution of 0.150 M KOH is titrated with 0.300 M HCl. Calculate the pH of the solution after the addition of the following amounts of HCl.

d) 18.0 mL HCl

e) 24.0 mL HCl

g) 26.0 mL HCl

h) 29.0 mL HCl

Answer 1

To calculate the pH of the solution after the addition of each amount of HCl, we will use the concept of stoichiometry and neutralization reaction between KOH and HCl. KOH (potassium hydroxide) is a base, and HCl (hydrochloric acid) is an acid.

The balanced chemical equation for the reaction is:

KOH + HCl → KCl + H2O

The reaction shows that 1 mole of KOH reacts with 1 mole of HCl to produce 1 mole of KCl and 1 mole of water.

Given data:

• Volume of KOH solution (V1) = 50.0 mL = 0.050 L

• Concentration of KOH (C1) = 0.150 M

• Volume of HCl added (V2) in each case = {18.0 mL, 24.0 mL, 26.0 mL, 29.0 mL} = {0.018 L, 0.024 L, 0.026 L, 0.029 L}

• Concentration of HCl (C2) = 0.300 M

Step 1: Calculate moles of KOH (n1) in the initial solution:

n1 = C1 * V1 n1 = 0.150 M * 0.050 L n1 = 0.0075 moles

Step 2: Determine the limiting reagent:

Since the stoichiometry of the reaction is 1:1, the limiting reagent will be the one that gets consumed completely. It will be the reactant with fewer moles.

Let's compare the moles of KOH (n1) and moles of HCl (n2) in each case:

For d) 18.0 mL HCl: n2 = C2 * V2 = 0.300 M * 0.018 L = 0.0054 moles KOH is the limiting reagent.

For e) 24.0 mL HCl: n2 = C2 * V2 = 0.300 M * 0.024 L = 0.0072 moles KOH is the limiting reagent.

For g) 26.0 mL HCl: n2 = C2 * V2 = 0.300 M * 0.026 L = 0.0078 moles KOH is the limiting reagent.

For h) 29.0 mL HCl: n2 = C2 * V2 = 0.300 M * 0.029 L = 0.0087 moles KOH is the limiting reagent.

Step 3: Calculate moles of KOH remaining after the reaction (n1_remaining):

n1_remaining = n1 - n2 (moles of KOH initially - moles of KOH reacted)

Step 4: Calculate the concentration of KOH remaining (C1_remaining) in the total volume after the addition of HCl:

V_total = V1 + V2 (initial volume + volume of HCl added)

For each case, calculate C1_remaining and then the pH:

a) For 18.0 mL HCl:

n1_remaining = 0.0075 moles - 0.0054 moles = 0.0021 moles V_total = 0.050 L + 0.018 L = 0.068 L C1_remaining = n1_remaining / V_total = 0.0021 moles / 0.068 L ≈ 0.0309 M

pH = -log(C1_remaining) = -log(0.0309) ≈ 1.51

b) For 24.0 mL HCl:

n1_remaining = 0.0075 moles - 0.0072 moles = 0.0003 moles V_total = 0.050 L + 0.024 L = 0.074 L C1_remaining = n1_remaining / V_total = 0.0003 moles / 0.074 L ≈ 0.0041 M

pH = -log(C1_remaining) = -log(0.0041) ≈ 2.39

c) For 26.0 mL HCl:

n1_remaining = 0.0075 moles - 0.0078 moles = -0.0003 moles (negligible) V_total = 0.050 L + 0.026 L = 0.076 L C1_remaining = n1_remaining / V_total = -0.0003 moles / 0.076 L ≈ -0.0039 M (negligible)

Since the moles of KOH remaining are negligible, the pH will be determined mainly by the excess HCl. The pH in this case is practically the same as when only HCl is present in the solution.

d) For 29.0 mL HCl:

n1_remaining = 0.0075 moles - 0.0087 moles = -0.0012 moles (negligible) V_total = 0.050 L + 0.029 L = 0.079 L C1_remaining = n1_remaining / V_total = -0.0012 moles / 0.079 L ≈ -0.0152 M (negligible)

As in the previous case, the moles of KOH remaining are negligible, and the pH will be determined mainly by the excess HCl. The pH is practically the same as when only HCl is present in the solution.

Note: In cases c) and d), where the moles of KOH remaining are negligible, the pH will be mainly determined by the excess HCl concentration. The pH calculation would involve finding the concentration of the excess HCl and using the appropriate formula to calculate the pH based on its concentration. However, since the concentration of KOH remaining is very small (close to zero), the contribution to the pH from the remaining KOH is negligible, and the pH is mainly determined by the excess HCl concentration.

Answer 2