已知100℃时,0.01mol-L -1 NaHSO 4 溶液中水电离的c(H + )=10 -10 mol-L -1 ,该温度下将pH=8的Ba(

2025-04-06 21:51:22
推荐回答(1个)
回答1:

硫酸氢钠溶液中水电离出的氢离子浓度等于溶液中氢氧根离子浓度,0.01mol?L -1  NaHSO 4 溶液中C(H + )=0.01mol/L,水电离的c(H + )=c(OH - )=10 -10  mol?L -1 ,所以K w =10 -12
该温度下,氢氧化钡溶液中c(OH - )=
1 0 -12
1 0 -8
mol/L=10 -4 mol/L,混合溶液呈碱性,
溶液中c(OH - )=
1 0 -12
1 0 -7
mol/L=10 -5 mol/L,
1 0 -4 × V 1 -1 0 -5 V 2
V 1 + V 2
=1 0 -5

所以V 1 :V 2 =2:9,故选A.