求化学竞赛题的详解设硫酸为二元强酸,现用0.01mol\LH2SO4溶液滴定0.01mol\LNaoH溶液,中和后加水至100mL.若滴定终点判断有误差:1.多加了1 滴H2SO4溶液;2.少加了1滴H2SO4溶液(1滴溶液的体积约为0.05m
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![求化学竞赛题的详解设硫酸为二元强酸,现用0.01mol\LH2SO4溶液滴定0.01mol\LNaoH溶液,中和后加水至100mL.若滴定终点判断有误差:1.多加了1 滴H2SO4溶液;2.少加了1滴H2SO4溶液(1滴溶液的体积约为0.05m](/uploads/image/z/13781718-54-8.jpg?t=%E6%B1%82%E5%8C%96%E5%AD%A6%E7%AB%9E%E8%B5%9B%E9%A2%98%E7%9A%84%E8%AF%A6%E8%A7%A3%E8%AE%BE%E7%A1%AB%E9%85%B8%E4%B8%BA%E4%BA%8C%E5%85%83%E5%BC%BA%E9%85%B8%2C%E7%8E%B0%E7%94%A80.01mol%5CLH2SO4%E6%BA%B6%E6%B6%B2%E6%BB%B4%E5%AE%9A0.01mol%5CLNaoH%E6%BA%B6%E6%B6%B2%2C%E4%B8%AD%E5%92%8C%E5%90%8E%E5%8A%A0%E6%B0%B4%E8%87%B3100mL.%E8%8B%A5%E6%BB%B4%E5%AE%9A%E7%BB%88%E7%82%B9%E5%88%A4%E6%96%AD%E6%9C%89%E8%AF%AF%E5%B7%AE%EF%BC%9A1.%E5%A4%9A%E5%8A%A0%E4%BA%861+%E6%BB%B4H2SO4%E6%BA%B6%E6%B6%B2%EF%BC%9B2.%E5%B0%91%E5%8A%A0%E4%BA%861%E6%BB%B4H2SO4%E6%BA%B6%E6%B6%B2%EF%BC%881%E6%BB%B4%E6%BA%B6%E6%B6%B2%E7%9A%84%E4%BD%93%E7%A7%AF%E7%BA%A6%E4%B8%BA0.05m)
求化学竞赛题的详解设硫酸为二元强酸,现用0.01mol\LH2SO4溶液滴定0.01mol\LNaoH溶液,中和后加水至100mL.若滴定终点判断有误差:1.多加了1 滴H2SO4溶液;2.少加了1滴H2SO4溶液(1滴溶液的体积约为0.05m
求化学竞赛题的详解
设硫酸为二元强酸,现用0.01mol\LH2SO4溶液滴定0.01mol\LNaoH溶液,中和后加水至100mL.若滴定终点判断有误差:1.多加了1 滴H2SO4溶液;2.少加了1滴H2SO4溶液(1滴溶液的体积约为0.05mL),则1和2两种情况下溶液中氢离子之比的值是多少?
求化学竞赛题的详解设硫酸为二元强酸,现用0.01mol\LH2SO4溶液滴定0.01mol\LNaoH溶液,中和后加水至100mL.若滴定终点判断有误差:1.多加了1 滴H2SO4溶液;2.少加了1滴H2SO4溶液(1滴溶液的体积约为0.05m
:①多加了一滴H2SO4 (设一滴为0.05mL),
c(H+)=0.01mol/L*2*0.05mL/100mL=0.00001mol/L=10^-5mol/L
②少加了一滴H2SO4(设一滴为0.05mL),
c(OH-)=0.01mol/L*2*0.05/100=10^-5mol/L
c(H+)=10^-14/10^-5=10^-9mol/L
则①和②c(H+)的比值是
10^-5/10^-9=5*10^4
哎 楼上很对