Question

A beaker with 1.10×10² mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.20 mL of a 0.450 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Please show clear steps!

Answer 1

The idea here is that you need to use the Henderson-Hasselbalch equation to determine the ratio that exists between the concentration of the weak acid and of its conjugate base in the buffer solution.

Once you know that, you can use the total molarity of the acid and of the conjugate base to find the number of moles of these two chemical species present in the buffer.

So, the Henderson-Hasselbalch equation looks like this

In your case, you have acetic acid, CH3COOH, as the weak acid and the acetate anion, CH3COO−, as its conjugate base. The pKa of the acid is said to be equal to 4.74, which means that you have

pH = 4.74 + log([CH3COO−] / [CH3COOH])

The pH is equal to 5, and so

5.00 = 4.74 + log([CH3COO−] / [CH3COOH])

log([CH3COO−] / [CH3COOH]) = 0.26

This will be equivalent to

[CH3COO−] / [CH3COOH] = 1.8197

This means that your buffer contains 1.8197 times more conjugate base than weak acid

[CH3COO−] = 1.8197 × [CH3COOH]

Now, because both chemical species share the same volume, 110 mL, this can be rewritten as

which is

(1)

So, the buffer contains 1.8197 times more moles of acetate anions that of acetic acid.

Now, the total molarity of the buffer is said to be equal to 0.1 M. You thus have

[CH3COOH]+[CH3COO−]=0.10 M

Once again, use the volume of the buffer to write

This will be equivalent to

(2)

Use equations (1) and (2) to find how many moles of acetate ions you have in the buffer

1.8197⋅n_CH3COOH+n_CH3COOH=0.011

n_CH3COOH=0.011 / (1.8197+1) = 0.0039 moles CH₃COOH

This means that you have

n_CH3COO⁻ = 1.8197 x 0.0039 moles

n_CH3COO⁻ = 0.0071 moles CH₃COO⁻

Now, hydrochloric acid, HCl, will react with the acetate anions to form acetic acid and chloride anions, Cl−

HCl (aq) + CH₃COO⁻ (aq) → CH₃COOH (aq) + Cl⁻ (aq)

Notice that the reaction consumes hydrochloric acid and acetate ions in a 1:1 mole ratio, and produces acetic acid in a 1:1 mole ratio.

Use the molarity and volume of the hydrochloric acid solution to determine how many moles of strong acid you have

In your case, this gets you

The hydrochloric acid will be completely consumed by the reaction, and the resulting solution will contain

n_HCl = 0 moles→ completely consumed

n_CH3COO− = 0.0071 moles − 0.00324 moles = 0.00386 moles CH₃COO⁻

n_CH3COOH = 0.0039 moles + 0.00324 moles = 0.00714 moles CH₃COOH

The total volume of the solution will now be

V_total = 110 mL+ 7.20 mL= 117.20 mL

The concentrations of acetic acid and acetate ions will be

Use the Henderson-Hasselbalch equation to find the new pH of the solution

pH = 4.74 + log (0.0329 M / 0.0609 M)

pH = 4.472

Therefore, the pH of the solution decreased by