Question

What is the pH of an aqueous solution made by combining 35.81 mL of a 0.4364 M sodium acetate with 38.23 mL of a 0.3183 M solution of acetic acid to which 3.524 mL of a 0.0588 M solution of NaOH was added?

Answer 1

pH will be 4.86.

Step-by-step explanation:

On mixing sodium acetate (a salt of acetic acid) with acetic acid we get buffer solution.

Let us calculate the moles of each species taken:

Moles of sodium acetate = molarity X volume(L)

= 0.4364 X 35.81 X 10⁻³= 0.0156 mol

the moles of acetic acid = molarity X volume(L)

= 0.3183 X 38.23 X 10⁻³= 0.0121 mol

The moles of NaOH added = molarity X volume(L)

= 0.0588 X 3.524 X 10⁻³= 0.000207 mol

The base added will react with the acetic acid producing equal moles of sodium acetate

`CH_(3)COOH+NaOH-->CH_(3)COONa+H_(2)O`

The moles of acetic acid reacted = moles of base added = 0.000207 mol

The moles of acetic acid left = 0.0121-0.000207 = 0.0119 mol

The moles of sodium acetate formed = 0.000207 mol

The new moles of sodium acetate = 0.0156+0.000207=0.0158 mol

The pH of buffer is calculated using Henderson Hassalbalch's equation

pKa of acetic acid = 4.74

`pH=pKa+log([salt])/([acid])=4.74+log(0.0158)/(0.0119)=4.86`