1.0.001mol/L的醋酸溶液PH=_2.0.01mol/L的氨水溶液PH=__

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1.0.001mol/L的醋酸溶液PH=_______
2.0.01mol/L的氨水溶液PH=________

醋酸Ka = 1.8 × 10^(-5)
所以[H+] = √(0.001×1.8 × 10^(-5) )
pH = -lg[H+] = 3.87
氨Kb = 1.8 × 10^(-5)
所以[OH] = √(0.01×1.8 × 10^(-5) )
pH = -lg[H+] = 14 + lg[OH-] = 7.26