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What will the pH of 1.50 L of pure water water be if 2.0 mL of 4.0 M HCl is added? By how much has the pH changed? What will the pH of the solution in part b be if 2.0 mL of 4.0 M HCl is added? By how much has the pH changed?

Sagot :

Answer:

Part A

pH ≈ 2.273

Part B

ΔpH ≈ -4.726

Part C

pH ≈ 1.973

Part D

ΔpH ≈ -0.301

Explanation:

Part A

The pH of a solution is given by the negative concentration of hydrogen ions in the solution

2.0 mL = 0.002 L

The number of moles of HCl in 2.0 mL of 4.0 M HCl is given as follows;

1 Liter of 4.0 M HCl contains 4.0 moles of HCl

2.0 mL = 0.002 L 4.0 M HCl contains 0.002 L/(1 L) × 4.0 M = 0.008 moles of HCl

The concentration of 0.008 moles in 1.50 L is given as follows;

Concentration = The number of moles/(The volume in liters)

∴ The concentration of 0.008 moles in 1.50 L, C = 0.008 moles/(1.5 L + 0.002 L)

∴ The concentration of 0.008 moles in 1.50 L, C ≈ 0.00533 moles/liter = 0.00533 M HCl

Given that HCl is a strong acid, we have that HCl dissociates completely to give equal number of H⁺ and Cl⁻ ions;

The number of moles of [H⁺] in the solution = 0.00533 moles

The pH of the solution = -log[H⁺]

∴ pH = -log[5.33 × 10⁻³] ≈ 2.273

The pH of the 1.5 L of pure water will be approximately 2.273

Part B

The pH of the pure water has changed from neutral (pH = 7) tp pH = 2.273

The change in pH is ΔpH = 2.274 - 7 = -4.726

ΔpH ≈ -4.726

Part C

When 2.0 mL of the 4.0 M HCl is added, the solution above, we have;

C = (0.008 + 0.008)/(1.5 + 0.002 + 0.002) ≈ 1.06383 × 10⁻²

The concentration of the solution becomes, C ≈ 1.06383 × 10⁻² mole/liter

The pH becomes, pH = -log(1.06383 × 10⁻²) ≈ 1.973

Part D

The amount by which the pH has changed, ΔpH ≈ 1.973 - 2.274 = -0.301.

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