Calculate the pH of the resultant mixtures: a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2 c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH

To calculate the pH of the resultant mixtures, we need to first determine the moles of the acids and bases involved in each reaction. Then, we can find out whether there is an excess of acid or base after the reaction, which will help us calculate the pH. Let's break down each mixture step by step.

Mixture A: 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl

First, we calculate the moles of Ca(OH)2 and HCl:

• Moles of Ca(OH)2 = Volume (L) × Concentration (M) = 0.010 L × 0.2 mol/L = 0.002 mol
• Moles of HCl = Volume (L) × Concentration (M) = 0.025 L × 0.1 mol/L = 0.0025 mol

Next, we look at the reaction:

Ca(OH)2 + 2 HCl → CaCl2 + 2 H2O

From the balanced equation, 1 mole of Ca(OH)2 reacts with 2 moles of HCl. Therefore, 0.002 moles of Ca(OH)2 would require 0.004 moles of HCl. However, we only have 0.0025 moles of HCl available. This means HCl is the limiting reactant.

After the reaction, we have:

• Remaining HCl = 0.0025 mol - 0.002 mol = 0.0005 mol
• No Ca(OH)2 remains.

The total volume of the solution is 10 mL + 25 mL = 35 mL = 0.035 L. Now, we can find the concentration of the remaining HCl:

Concentration of HCl = Moles/Volume = 0.0005 mol / 0.035 L = 0.01429 M

Finally, we calculate the pH:

pH = -log[H+] = -log(0.01429) ≈ 1.84

Mixture B: 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2

First, we calculate the moles of H2SO4 and Ca(OH)2:

• Moles of H2SO4 = 0.010 L × 0.01 mol/L = 0.0001 mol
• Moles of Ca(OH)2 = 0.010 L × 0.01 mol/L = 0.0001 mol

The reaction is:

H2SO4 + Ca(OH)2 → CaSO4 + 2 H2O

From the balanced equation, 1 mole of H2SO4 reacts with 1 mole of Ca(OH)2. Since we have equal moles of both, they will completely neutralize each other.

After the reaction, there are no excess moles of acid or base, resulting in a neutral solution. Therefore, the pH of the mixture is:

pH = 7.00

Mixture C: 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH

Calculating the moles of H2SO4 and KOH:

• Moles of H2SO4 = 0.010 L × 0.1 mol/L = 0.001 mol
• Moles of KOH = 0.010 L × 0.1 mol/L = 0.001 mol

The reaction is:

H2SO4 + 2 KOH → K2SO4 + 2 H2O

From the balanced equation, 1 mole of H2SO4 reacts with 2 moles of KOH. Therefore, 0.001 moles of H2SO4 would require 0.002 moles of KOH. Since we only have 0.001 moles of KOH, KOH is the limiting reactant.

After the reaction, we have:

• Remaining H2SO4 = 0.001 mol - 0.0005 mol = 0.0005 mol
• No KOH remains.

The total volume of the solution is 10 mL + 10 mL = 20 mL = 0.020 L. Now, we can find the concentration of the remaining H2SO4:

Concentration of H2SO4 = 0.0005 mol / 0.020 L = 0.025 M

Finally, we calculate the pH:

pH = -log[H+] = -log(0.025) ≈ 1.60

Summary of pH Values

• Mixture A: pH ≈ 1.84
• Mixture B: pH = 7.00
• Mixture C: pH ≈ 1.60

These calculations illustrate how to determine the pH of mixtures involving strong acids and bases, taking into account their neutralization reactions and the resulting concentrations of the remaining species.