Calculate the expected pH values of the buffer systems from the experiments (a,

Question

Calculate the expected pH values of the buffer systems from the experiments (a, b, c, d) using the Henderson-HAsselbach equation, pH= pKa + log([A-]/[HA]). Use for pKa values carbonic acid = 6.37 and acetic acid=4.75

for A 5 ml 0.1 M CH3COOH + 5 mL 0.1 M CH3COO-NA+ pH 4.1 after addition of 0.5mL 0.1 M HCL pH is 4.2 after addition of 0.5 mL 0.1 M NaOH pH is 4.1

for B 1mL M CH3COOH + 10 mL 0.1 M CH3COO-Na+ pH is 5.2 after addition of 0.5 mL 0.1 M HCL pH is 5.1 after addition of 0.5 mL 0.1 NaOH pH is 6.0

for C 5mL 0.1 M H2CO3 + 5mL 0.1 M NaHCO3 pH is 9.2 after addition of 0.5 mL 0.1 M HCL pH is 8.9 after addition of 0.5 mL 0.1 M NaOH pH is 9.5

for D 1mL 0.1 M H2CO3 + 10mL 0.1 M NaHCO3 pH is 8.6 after addition of 0.5 mL o.1 M HCl pH is 7.5 after addition of 0.5 mL 0.1 M NaOH pH is 9.0

Explanation / Answer

1) A 5 ml 0.1 M CH3COOH + 5 mL 0.1 M CH3COO-NA+ pH 4.1 after addition of 0.5mL 0.1 M HCL pH is 4.2 after addition of 0.5 mL 0.1 M NaOH pH is 4.1

Solution: Moles of acetic acid = 5 X 0.1 = 0.5 millimoles ; Moles of salt = 5 X 0.1 = 0.5 millimoles of salt

Moles of HCl added = 0.5 X 0.1 = 0.05 millimoles of acid

Due to addition of HCl the moles of acid will increase by 0.05 and the moles of base will decrease by 0.05 moles

pH = pKa + log [salt] / [acid] = 4.75 + log [0.5-0.05] / [0.5+0.05] = 4.75 -0.087 = 4.663

Due to addition of NaOH the moles of acid will decrease by 0.05 and the moles of acid will increase by 0.05 moles

pH = pKa + log [salt] /[acid] = 4.75 + log [0.5+0.05] / [0.5-0.05] = 4.75 + 0.0871 = 4.837

2) for B 1mL M CH3COOH + 10 mL 0.1 M CH3COO-Na+ pH is 5.2 after addition of 0.5 mL 0.1 M HCL pH is 5.1 after addition of 0.5 mL 0.1 NaOH pH is 6.0

solution:

Moles of acetic acid = 10 X 0.1 = 1millimoles ; Moles of salt = 10 X 0.1 = 1 millimoles of salt

Moles of HCl added = 0.5 X 0.1 = 0.05 millimoles of acid

Due to addition of HCl the moles of acid will increase by 0.05 and the moles of base will decrease by 0.05 moles

pH = pKa + log [salt] / [acid] = 4.75 + log [1-0.05] / [1+0.05] = 4.75 -0.0434= 4.71

Due to addition of NaOH the moles of acid will decrease by 0.05 and the moles of acid will increase by 0.05 moles

pH = pKa + log [salt] /[acid] = 4.75 + log [1+0.05] / [1-0.05] = 4.75 + 0.0434 = 4.795

3) for C 5mL 0.1 M H2CO3 + 5mL 0.1 M NaHCO3 pH is 9.2 after addition of 0.5 mL 0.1 M HCL pH is 8.9 after addition of 0.5 mL 0.1 M NaOH pH is 9.5

Solution

Moles of carbonic acid = 5 X 0.1 = 0.5 millimoles ; Moles of salt = 5 X 0.1 = 0.5 millimoles of salt

Moles of HCl added = 0.5 X 0.1 = 0.05 millimoles of acid

Due to addition of HCl the moles of acid will increase by 0.05 and the moles of base will decrease by 0.05 moles

pH = pKa + log [salt] / [acid] = 6.37 + log [0.5-0.05] / [0.5+0.05] = 6.37 -0.087 = 6.283

Due to addition of NaOH the moles of acid will decrease by 0.05 and the moles of acid will increase by 0.05 moles

pH = pKa + log [salt] /[acid] = 6.37 + log [0.5+0.05] / [0.5-0.05] = 6.37 + 0.0871 = 6.457

4) for D 1mL 0.1 M H2CO3 + 10mL 0.1 M NaHCO3 pH is 8.6 after addition of 0.5 mL o.1 M HCl pH is 7.5 after addition of 0.5 mL 0.1 M NaOH pH is 9.0

Solution

Moles of carbonic acid = 10 X 0.1 = 1millimoles ; Moles of salt = 10 X 0.1 = 1 millimoles of salt

Moles of HCl added = 0.5 X 0.1 = 0.05 millimoles of acid

Due to addition of HCl the moles of acid will increase by 0.05 and the moles of base will decrease by 0.05 moles

pH = pKa + log [salt] / [acid] = 6.37 + log [1-0.05] / [1+0.05] = 6.37-0.0434= 6.327

Due to addition of NaOH the moles of acid will decrease by 0.05 and the moles of acid will increase by 0.05 moles

pH = pKa + log [salt] /[acid] = 6.37 + log [1+0.05] / [1-0.05] = 6.37 + 0.0434 = 6.413

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