Question

The Ka of bromoacetic acid is 2.00 ×10–3. What masses of bromoacetic acid (CH2BrCOOH) and sodium bromoacetate (CH2BrCOONa) are needed to prepare 1.00 L of pH = 2.100 buffer if the total concentration of the two components is 0.200 M? g of bromoacetic acid g of sodium bromoacetate

Answer 1

g of bromoacetic acid = 22.2 grams

g of sodium bromoacetate = 6.47 grams

Explanation

Total concentration = 0.200 M

[CH₂BrCOOH] + [CH₂BrCOONa] = 0.200 M     ...(1)

Ka = 2.00 x 10^-3

pKa = -log(Ka)

pKa = -log(2.00 x 10^-3)

pKa = 2.70

According to Henderson - Hasselbalch equation,

pH = pK_a + log([conjugate base] / [weak acid])

2.100 = 2.70 + log([CH₂BrCOONa] / [CH₂BrCOOH])

log([CH₂BrCOONa] / [CH₂BrCOOH]) = 2.100 - 2.70

log([CH₂BrCOONa] / [CH₂BrCOOH]) = -0.60

[CH₂BrCOONa] / [CH₂BrCOOH] = 10^-0.60

[CH₂BrCOONa] / [CH₂BrCOOH] = 0.252      ...(2)

Solving equations (1) and (2)

[CH₂BrCOOH] = 0.160 M

[CH₂BrCOONa] = 0.040 M

moles CH₂BrCOONa = (molarity CH₂BrCOONa) * (volume of buffer in Liter)

moles CH₂BrCOONa = (0.040 M) * (1.00 L)

moles CH₂BrCOONa = 0.040 mol

mass CH₂BrCOONa = (moles CH₂BrCOONa) * (molar mass CH₂BrCOONa)

mass CH₂BrCOONa = (0.040 mol) * (160.93 g/mol)

mass CH₂BrCOONa = 6.47 g

moles CH₂BrCOOH = (molarity CH₂BrCOOH) * (volume of buffer in Liter)

moles CH₂BrCOOH = (0.160 M) * (1.00 L)

moles CH₂BrCOOH = 0.160 mol

mass CH₂BrCOOH = (moles CH₂BrCOOH) * (molar mass CH₂BrCOOH)

mass CH₂BrCOOH = (0.160 mol) * (138.95 g/mol)

mass CH₂BrCOONa = 22.2 g