Question

How much 5.5 M NaOH must be added to 590 mL of a buffer that is 0.0195M acetic acid and 0.0250M sodium acetate to raise the pH to 5.75?

Answer 1

Given the volume of the buffer solution, V = 590 mL = 590 mL x (1L / 1000 mL) = 0.590 L

Moles of acetic acid (CH3COOH) in the buffer = MxV(L) = 0.0195 M x 0.590L = 0.011505 mol

moles of  sodium acetate(CH3COONa) in the buffer =  MxV(L) = 0.0250 M x 0.590L = 0.01475 mol

Let the volume of 5.5M NaOH added be V L.

Hence moles of NaOH added = MxV = 5.5V mol

NaOH reacts with acetic acid (CH3COOH) to form sodium acetate(CH3COONa). The balanced chemical reaction is

--------------- CH3COOH + NaOH ------ > CH3COONa + H2O

Initial mol: 0.011505, --- 5.5V ----------- 0.01475

---Change: - 5.5V, -------- - 5.5V, --------- + 5.5V

eqm.mol: (0.011505 - 5.5V), 0, -----------(0.01475 + 5.5V)

Hence after neutralization reaction,

[CH3COOH] = (0.011505 - 5.5V) / (0.590+V)

[CH3COONa] = (0.01475 + 5.5V) / (0.590+V)

Applying Hendersen equation

pH = pKa + log[CH3COONa] / [CH3COOH]

=> 5.75 = 4.76 + log(0.01475 + 5.5V) / (0.011505 - 5.5V)

=> log(0.01475 + 5.5V) / (0.011505 - 5.5V) = 0.99

=> (0.01475 + 5.5V) / (0.011505 - 5.5V) = 9.7724

=> V = 0.00165 L = 1.65 mL (answer)