电荷守恒 c(CH3COO-) + c(OH-) = c(Na+) + c(H+) ①
因为溶液为中性,所以c(OH-) = c(H+) = 1×10^(-7) mol/L ②
由①②得,c(CH3COO-) = c(Na+) = 0.001/2 = 5×10^(-4) mol/L(等体积混合后,Na+浓度减半)
物料守恒 c(CH3COO-) + c(CH3COOH) = m/2 mol/L
所以 c(CH3COOH) = m/2 mol/L - c(CH3COO-) = [ m/2 - 5×10^(-4) ] mol/L
CH3COOH ===可逆=== CH3COO- + H+
m/2 - 5×10^(-4) 5×10^(-4) 1×10^(-7)
K = c(CH3COO-) c(H+) / c(CH3COOH)
= 5×10^(-11) / [ m/2 - 5×10^(-4) ] = 1×10^(-10) / ( m - 0.001 )