Question

A buffer solution (pH 4.74) contains acetic acid (0.05 mol/L) and sodium acetate (0.05 mol/L) i.e....

A buffer solution (pH 4.74) contains acetic acid (0.05 mol/L) and sodium acetate (0.05 mol/L) i.e. it is a 0.1M acetate buffer. Calculate the pH after addition of 2 mL of 0.025M hydrochloric acid to 10 mL of the buffer.​

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Homework Answers

Answer #1

4.65

Explanation

No of mole of Hydrochloric acid added = (0.025mol/1000ml)×2ml = 0.00005mol

No of mole of CH3COOH = (0.05mol/1000ml)×10ml = 0.0005mol

No of mole of CH3COO- = (0.05mol/1000ml)×10ml = 0.0005mol

HCl react with CH3COO-

HCl + CH3COO- ------> CH3COOH + Cl-

this is 1:1 reaction

Therefore, after HCl addition

No of mole of CH3COOH = 0.00050 + 0.00005 = 0.00050

No of mole of CH3COO- = 0.00050 - 0.00005 = 0.00045

total volume = 12ml

[ CH3COOH ] = (0.00055mol/12ml)×1000ml = 0.0458M

[ CH3COO- ] = (0.00045mol/12ml) ×1000ml = 0.0375M

Handerson equation

pH = pKa + log ([A-]/[HA])

= 4.74 + log( 0.0375/0.0458)

= 4.74 - 0.09

= 4.65

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