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A 200.0 mL buffer solution is 0.260 M in acetic acid and 0.260 M in sodium...

A 200.0 mL buffer solution is 0.260 M in acetic acid and 0.260 M in sodium acetate.

1) What is the initial pH of this solution?

2) What is the pH after addition of 0.0100 mol of HCl?

3) What is the pH after addition of 0.0100 mol of NaOH?

Expert Solution

1) The pH of buffer can be calculated by using Hendersen Hassalbalch equaiton

pH = pKa + log[ salt] / [acid]

pKa of acetic acid =4.75

pH = 4.75 + log [0.26 / .26]

pH = 4.75

2) After addition of HCl it will react with salt to give acetic acid and will consume the salt

Moles of acid added = 0.01

CH3COONa + HCl --> CH3COOH + NaCl

Moles of Acetic acid formed = 0.01

Initial moles of acetic acid = concentration X volume = 0.26 x 0.2 = 0.052

Final moles of acetic acid = 0.052 + 0.01 = 0.062

Moles of sodium acetate consumed = 0.01

Initial moles of sodium acetate = 0.26 x 0.2 = 0.052

Final moles of sodium acetate = 0.052 - 0.01 = 0.042

New pH will be

pH = pKa + log [salt] / [acid]

pH = 4.75 + log [0.042 / 0.062] = 4.75 -0.169 = 4.58

3) when NaOH is added it reacts with CHeCOOH to give sodium acetate

NaOH + CH3COOH --> CH3COONa + H2O

Moles of NaOH added = 0.01

Moles of Acetic acid consumed = 0.01

Initial moles of acetic acid = concentration X volume = 0.26 x 0.2 = 0.052

Final moles of acetic acid = 0.052 - 0.01 = 0.042

Moles of sodium acetate formed = 0.01

Initial moles of sodium acetate = 0.26 x 0.2 = 0.052

Final moles of sodium acetate = 0.052 + 0.01 = 0.062

New pH will be

pH = pKa + log [salt] / [acid]

pH = 4.75 + log [0.062 / 0.042] = 4.75 + 0.169 = 4.92

queen_honey_blossom answered 1 week ago

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