配置pH=9的缓冲溶液用0.2mol/l的氨水500,问1mol/的HCL体积多少
人气：374 ℃ 时间：2019-11-25 12:35:15
解答
一般缓冲溶液pH = p Ka +- 1 ,Kb(NH3·H2O)=10^(-4.74) Ka(NH4+)=10^(-9.26) pKa=9.26,显然可用氨水和氯化铵配制pH=9的缓冲溶液从Kb(NH3·H2O)= [NH4+]·[OH-]/[NH3·H2O]约=C（NH4+）·[OH-]/C（NH3·H2O）=n（NH4+...
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