室温下将PH=12的NaOH溶液aL与ph=1的H2SO4溶液bL混合(1)PH=12的NaOH溶液中水电离出OH-浓度为 常温时,将pH=3的H2SO4溶液和pH=12的NaOH溶液...

Kw=10^-12\uff0c\u6df7\u5408ph=11\u7684naoh\u6eb6\u6db2aL\u548cph=1\u7684h2so4\u6eb6\u6db2bL\u540eph=2,a:

NaoH\u4e2d\u7684H+\u548cH2SO4\u4e2d\u7684oH-\u53ef\u5ffd\u7565\u4e0d\u8ba1
\u7531\u9898\u53ef\u5f97\uff1aNaoH\u548cH2SO4\u90fd\u4e3a10mol/L
\u5219\uff1a10mol/L*b -10mol/L*a
\u2014\u2014\u2014\u2014\u2014\u2014-\u2014\u2014\u2014\u2014=0.01mol/L \u5316\u7b80\u5f97 a:b=9:11
a+b

pH=3\u7684H2SO4\u6eb6\u6db2c\uff08H+\uff09=10-3 mol/L\uff0cpH=12\u7684NaOH\u6eb6\u6db2\u4e2dc\uff08OH-\uff09=10-2mol/L\uff0c\u6df7\u5408\u6eb6\u6db2\u4e2dc\uff08OH-\uff09=10-4mol/L=c(OH?)V(NaOH)?c(H+)\uff0eV(H2SO4)V(H2SO4)+V(NaOH)=0.01\u00d7V(NaOH)?0.001\u00d7V(H2SO4)V(H2SO4)+V(NaOH)\uff0cV\uff08H2SO4\uff09\uff1aV\uff08NaOH\uff09=99\uff1a11=9\uff1a1\uff0c\u6545\u9009B\uff0e

解:1、pH=12的NaOH溶液中水电离出C(H^+)=1*10^-12mol/L,
水电离出的C(OH^-)=水电离出C(H^+)=1*10^-12mol/L
2、pH=12的NaOH溶液中C(H^+)=1*10^-12mol/L
则C(OH)^-)=Kw/C(H^+)=1*10^-2mol/L
pH=1的H2SO4溶液中C(H^+)=1*10^-1mol/L
若所得的混合液为中性,说明氢离子与氢氧根离子等物质的量中和,即:
C1V1=C2V2
1*10^-2mol/L*AL=1*10^-1mol/L*BL
A:B=10:1
3、若所得的混合液的PH=2,说明酸过量,即:
(C2V2-C1V1)/(V1+V2)=10^-2mol/L
(10^-1mol/L*BL-10^-2mol/L*AL)/(A+B)=10^-2mol/L
A:B=9:2
希望我的回答能对你的学习有帮助!

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