Question

Calculate the change in pH that occurs when 0.00100 mol of gaseous HCl is added to a buffer solution that is prepared by dissolving:

A) 4.92g of sodium acetate (molar mass = 82.03g/mol) in 250 mL of 0.150 mol/L of acetic acid solution?

B) 0.492g of sodium acetate (molar mass = 82.03g/mol) in 250 mL of 0.0150 mol/L of acetic acid solution?

Assume no change in volume upon addition of either the sodium acetate or HCl. (Ka for acetic acid = 1.8x10^-5).

Answer 1

A) 4.92g of sodium acetate (molar mass = 82.03g/mol) in 250 mL of 0.150 mol/L of acetic acid solution?

1) Calculation of pH of buffer before addition of HCl gas.

Molar mass of Sodium acetate AcONa = 82.03 g/mole

Mass of AcONa added = 4.92 g.

# of moles of AcONa = Given mass / Molar mass = 4.92 / 82.03 = 0.06 mole.

Volume of solution = 250 mL = 0.250 L

Hence, [AcONa] = # of moles of AcONa / Volume of solution in L = 0.06 / 0.250 = 0.24 M/L.

pH of buffer is given by Henderson Hasselbalch equation as,

pH = pKa + log([Conjugate base]/[Acid])

For AcOH-AcONa buffer,

pH = pKa + log([AcONa]/[AcOH]) -----------(1)

Ka of AcOH = 1.8 x 10^-5.

pKa of AcOH = -log(1.8 x 10^-5) = 4.745

[AcOH] = 0.15 M, [AcONa] = 0.24 M.

Using these values in eq.(1) we get,

pH = 4.745 + log (0.24/0.15)

pH = 4.745 + 0.204

pH = 4.949

2) Calculation of pH of buffer on addition of 0.001 moles of HCl gas.

[HCl] = # of moles of HCl / Volume of solution in L = 0.001/0.250 = 0.004 M.

On addition of 0.004 M HCl, [AcOH] will increase and of [AcO-] decrease.

New [AcOH] = 0.150 + 0.004 = 0.154 M

new [AcONa] = 0.240 0.004 = 0.236 M

Using these new concentrations in eq.(1) we get,

pH = 4.745 + log (0.236/0.154)

pH = 4.745 + 0.185

pH = 4.930

Change in pH = 4.949 - 4.930 = 0.019 units.

pH of original buffer will decrease by 0.019 units.

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B)

1) Calculation of pH of buffer before addition of HCl gas.

Molar mass of Sodium acetate AcONa = 82.03 g/mole

Mass of AcONa added = 0.492 g.

# of moles of AcONa = Given mass / Molar mass = 0.492 / 82.03 = 0.006 mole.

Volume of solution = 250 mL = 0.250 L

Hence, [AcONa] = # of moles of AcONa / Volume of solution in L = 0.006 / 0.250 = 0.024 M/L.

pH of buffer is given by Henderson Hasselbalch equation as,

pH = pKa + log([Conjugate base]/[Acid])

For AcOH-AcONa buffer,

pH = pKa + log([AcONa]/[AcOH]) -----------(1)

Ka of AcOH = 1.8 x 10^-5.

pKa of AcOH = -log(1.8 x 10^-5) = 4.745

[AcOH] = 0.015 M, [AcONa] = 0.024 M.

Using these values in eq.(1) we get,

pH = 4.745 + log (0.024/0.015)

pH = 4.745 + 0.204

pH = 4.949

2) Calculation of pH of buffer on addition of 0.001 moles of HCl gas.

[HCl] = # of moles of HCl / Volume of solution in L = 0.001/0.250 = 0.004 M.

On addition of 0.004 M HCl, [AcOH] will increase and of [AcO-] decrease.

New [AcOH] = 0.015 + 0.004 = 0.019 M

new [AcONa] = 0.024 -0.004 = 0.020 M

Using these new concentrations in eq.(1) we get,

pH = 4.745 + log (0.020/0.019)

pH = 4.745 + 0.022

pH = 4.767

Change in pH = 4.949 - 4.767 = 0.182 units.

pH of original buffer will decrease by 0.182 units.

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