A beaker with 110 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.80 mL of a 0.400 MM HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
Answer
∆pH = -0.58
Explanation
Henderson - Hasselbalch equation is
pH = pKa + log([A-]/[HA])
5.00 = 4.760 + ([CH3COO-]/[CH3COOH])
log[CH3COO-] /[CH3COOH] = 0.24
[CH3COO-] / [CH3COOH]= 1.738
[CH3COO-] = 1.738[CH3COOH]
[CH3COO-] + [CH3COOH] = 0.100M
(1.738 × [CH3COOH]) + [CH3COOH] = 0.100M
2.738 × [CH3COOH] = 0.100M
[CH3COOH] = 0.03652M
[CH3COO-] = 0.100M - 0.03652M = 0.06348M
moles of CH3COOH = ( 0.03652mol/1000ml) × 110ml = 0.0040172mol
moles of CH3COO- = ( 0.06348mol/1000ml) × 1!0ml = 0.0069828mol
HCl react with conjucate base CH3COO-
CH3COO- + H+ --------> CH3COOH
moles of HCl added = ( 0.400mol/1000ml) × 8.80mln= 0.00352 mol
after addition of HCl
moles of CH3COOH = 0.0040172mol + 0.00352mol = 0.0075372mol
moles of CH3COO- = 0.0069828mol - 0.00352mol = 0.0034627mol
Total volume = 110ml + 8.80ml = 118.80ml
[CH3COOH] = (0.0075372mol/118.80ml) ×1000ml = 0.06344M
[CH3COO-] = (0.0034627mol/118.80ml) × 1000ml = 0.02915M
Applying Henderson - Hasselbalch equation
pH = 4.76 + log( 0.02915M/0.06344M)
pH = 4.76 - 0.34
pH = 4.42
∆pH = 4.42 - 5.00
∆pH = - 0.58
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