Question

What is the pH when 32.5 mL of 0.168 M Acetic acid reacts with 25.1 mL of 0.111 M potassium hydroxide at 25 deg. C? K_a = 1.8x10^-5

Answer 1

Ans:-

Moles of acetic acid = 0.168× 32.5 × 10^(-3) = 0.00546 moles of NaOH = 0.111× 25.1 × 10^(-3) = 0.0027861

CH3COOH + NaOH ------> CH3COONa + H20   

0.0027861 mol of NaOH will neutralise 0.0027861 mol of acetic acid and same amount of CH3COONa will be formed.

Total volume = 32.5 + 25.1 = 57.6ml

Moles of CH3COONa formed = 0.0027861

[CH3COONa] = 0.0027861× 1000/57.6 = 0.0484 M

Moles of CH3COOH left = 0.0026739

[CH3COOH]= 0.0026739×1000/57.6= 0.0464 M

pH = pKa+ log (CH3COONa / CH3COOH )

pH = - log ( 1.8 ×10^-5) + log(0.0484/0.0464)

pH=