When PH = -㏒[H3O+] and we have PH = 5.061 by substitution: ∴ [ H3O+] = 10^-5.061 = 8.7x10^-6 when we have Ka = [CH3COO-][H3O+] / [CH3COOH] we have Ka = 1.8x10^-5 & [H3O+] = 8.7x10^-6m & [CH3COOH] = 0.5 m by substitution in Ka formula: 1.8x10^-5 = [CH3COO-]*(8.7x10^-6) / 0.5 ∴[CH3COO-] = 1.034 ∴we need to add 1.034 mol of sodium acetate