Question

If 0.0080 moles of solid sodium hydroxide is dissolved into a 1.0 L 0.020 M acetic acid solution, what is the final pH of this solution? (pKa of acetic acid is 4.80 and assume no volume change during the dissolving process). Can you also explain the logic behind the drop in pH

Answer 1

Given: 1.0 L of 0.020 M acetic acid

mol of acetic acid=0.020 mol/L*1L=0.020 mol

NaOH +CH3COOH --->CH3COONa +H2O

Thus, 0.0080 mol of acetic acid is neutralized by NaOH

Remaining mol of CH3COOH=0.020mol-0.0080mol=0.012mol

mol of salt of acid (CH3COONa)=0.0080mol

pH=pka+log [CH3COONa]/[CH3COOH] {henderson-hasselbach equation)

=4.80+log (0.008mol/0.012mol)=4.62

pH=4.62

pH drops as conjugate base of acetic acid (CH3COO-) is formed on neutralization by NaOH

base lowers the pH,by increasing OH- concentration in water on its hydrolysis,

CH3COO^- +H2O ----->CH3COOH +OH^-

^pH=-log[OH-]

^pH=14-pOH