根据题意,完成下列问题.(1)常温下,将1mL pH=1的H2SO4溶液加水稀释到100mL,稀释后的溶液pH=______.

2025-05-14 00:30:19
推荐回答(1个)
回答1:

(1)1mL pH=1的H2SO4溶液加水稀释到100mL,c(H+)由0.1mol/L变为0.001mol/L,则pH=-lg0.001=3,
故答案为:3;
(2)常温下,0.01mol?L-1的NaOH溶液的pH为12,而某温度时测得0.01mol?L-1的NaOH溶液的pH为11,则KW=0.01×10-11=1.0×10-13,温度高于25℃,
故答案为:1.0×10-13
(3)pH=5的H2SO4的溶液中由水电离出的H+浓度为c1=

10?14
10?5
=10-9mol/L,pH=9的Ba(OH)2溶液中由水电离出的H+浓度为c2=10-9mol/L,则
c1
c2
=1,故答案为:1;
(4)若所得混合溶液呈中性,aL×0.1mol/L=bL×0.001mol/L,解得a:b=1:100;若所得混合溶液pH=12,则碱过量,所以
a×0.1?b×0.001
a+b
=0.01mol/L,解得a:b=11:90,
故答案为:1:100;11:90.