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Nata [24]
3 years ago
8

A 1.00 L buffer solution is 0.112 M in acetic acid and 0.112 M in sodium acetate. Acetic acid has a pKa of 4.74. What is the pH

change of this solution upon addition of 0.1 mol of HCl?
Chemistry
1 answer:
topjm [15]3 years ago
6 0

Answer:

ΔpH = 1.25

Explanation:

Using Henderson-Hasselbalch formula:

pH = pka + log [CH₃COONa] / [CH₃COOH]

pH = 4.74 + log [0.112] / [0.112]

<em>pH = 4.74</em>

The reaction of sodium acetate (CH₃COONa) with HCl is:

CH₃COONa + HCl → CH₃COOH + NaCl

<em>Producing acetic acid, </em>CH₃COOH.

If 0.1mol of HCl reacts the final moles of CH₃COONa are:

0.112mol - 0,1 mol = 0.012mol

Moles of acetic acid are:

0.112mol + 0,1 mol = 0.212mol

Using Henderson-Hasselbalch formula:

pH = pka + log [CH₃COONa] / [CH₃COOH]

pH = 4.74 + log [0.012] / [0.212]

<em>pH = 3.49</em>

<em></em>

Change in pH, ΔpH = 4.74 - 3.49 =<em> 1.25</em>

I hope it helps!

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