Question

If 10 mL of 1 M NaOH are added to one liter of a buffer that is 0.3 M acetic acid and 0.2 M sodium acetate (Na+CH3COO-), how much does the pH change? The pKa for acetic acid is 4.76.

a: 0.05 units

b: 1.2 units

c: 0.32 units

d: 3 units

Answer 1

Solution :-

Lets first calculate the moles of each species

Moles of acetic acid = 0.3 mol per L * 1 L = 0.3 mol

Moles of acetate = 0.2 mol per L * 1 L = 0.2 mol

Moles of NaOH = 1 mol per L * 0.01 L = 0.01 mol

When the NaOH is added to buffer then it will react with acetic acid and forms acetat

So the moles of acetic acid remain = 0.3 mol – 0.01 mol = 0.29 mol

Moles of acetate in the solution = 0.2 mol + 0.01 mol = 0.021 mol

Now lets calculate the pH of the solution

pH= pka + log ([acetate]/[acetic acid])

pH= 4.76 + log [0.21/0.29]

pH= 4.62

lets calculate the initial pH

pH= pka + log ([acetate]/[acetic acid])

pH= 4.76 + log [0.2/0.3]

pH= 4.58

so the change in pH = 4.62 – 4.58 = 0.04

so the closest answer is the 0.05 unit

so the answer is option a that is 0.05 units