Suppose 50.00 mL of 0.300 M HCl is added to an acetate buffer prepared by dissolving...

Suppose 50.00 mL of 0.300 M HCl is added to an acetate buffer prepared by dissolving 0.100 mol of acetic acid and 0.110 mol of sodium acetate in 0.100 L of solution. What are the initial and final pH values? What would be the pH if the same amount of HCl solution were added to 170 mL of pure water?

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Expert Solution

pK_a of acetic acid (denote as HA) is 4.75. We have 0.100 mole HA and 0.110 mole sodium acetate (NaA) in 0.100 L solution to start with.

[HA] = (moles of HA)/(volume of solution in L) = (0.100 mole)/(0.100 L) = 1.000 mol/L = 1.000 M.

[NaA] = (moles of NaA)/(volume of solution in L) = (0.110 mole)/(0.100 L) = 1.100 mol/L = 1.100 M.

Use the Henderson-Hasslebach equation to compute the initial pH of the solution.

pH = pK_a + log [NaA]/[HA] = 4.75 + log (1.100 M)/(1.000 M) = 4.75 + log (1.100) = 4.75 + 0.04139 = 4.79139 ≈ 4.79 (ans).

Now 50.00 mL of 0.300 M HCl is added; moles of HCl added = (50.00 mL)*(1 L/1000 mL)*(0.300 M)*(1 mol/L/1 M) = 0.015 mole.

HCl reacts with NaA as below to give HA.

NaA (aq) + HCl (aq) -------> NaCl (aq) + HA (aq)

As per the stoichiometric equation,

1 mole HCl = 1 mole NaA = 1 mole HA.

Therefore, moles HCl added = moles NaA neutralized = moles HA formed = 0.015 mole.

Moles NaA retained at equilibrium = (0.110 – 0.015) mole = 0.095 mole.

Moles HA at equilibrium = (0.100 + 0.015) mole = 0.115 mole.

The volume of the solution is constant for both HA and NaA; hence, we can express the equilibrium concentrations in terms of number of moles of HA and NaA.

Again, use the Henderson-Hasslebach equation as below.

pH = pK_a + log [NaA]/[HA] = 4.75 + log (0.095 mole)/(0.115 mole) = 4.75 + log (0.826087) = 4.75 + (-0.08297) = 4.66703 ≈ 4.67 (ans).

The pH of water is 7.00 and water is neutral; consequently, there are no weakly acidic or basic species in water and the pH is simply the pH of HCl. However, note that the HCl has been diluted due to addition to 170.00 mL water.

Moles of HCl = 0.015 mole; total volume of the solution = (50.00 + 170.00) mL = 220.00 mL = (220.00 mL)*(1 L/1000 mL) = 0.220 L.

[HCl] = [H⁺] = (0.015 mole)/(0.220 L) = 0.06818 mole/L = 0.06818 M.

pH = -log [H⁺] = -log (0.06818) = 1.1663 ≈ 1.17 (ans).

queen_honey_blossom answered 1 year ago

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