例10计算0.033moL1Na2HPO4的pH pKa1 -pK Q1 a3' 216,721,1232 解:Ka3C=101232×0033=101380≈K c/Ka2=0.03310721=10573>>20 故 K C+K H a3 =10 9.66 cK a2 pH=966 (如用最简式,pH=9.77,E=22%) 1515 例10 计算 0.033mol·L-1 Na2HPO4 的pH. pKa1 ~ pKa3: 2.16, 7.21, 12.32 解: Ka3c = 10-12.32×0.033 = 10-13.80 ≈ Kw c/Ka2 = 0.033/10-7.21 = 105.73 >> 20 故: a w 3 9.66 a2 [H ] 10 K c + K = = c/K + - (如用最简式, pH = 9.77, Er = 22%) pH = 9.66