1)將0.05mol/l鹽酸與未知濃度的NaOH溶液以1:2的體積比混合,所得溶液pH=2
10^(-2)mol/L= [0.05mol/L ×V - c(OH-)×2V]/3V
c(OH-)=0.01mol/L
2)用上述NaOH與某一元弱酸HA 20ml反應(yīng),恰好中和時(shí)消耗NaOH溶液10ml,所得溶液PH=10
此時(shí)溶液 c(Na+)= 0.01 mol/L×10mL÷30mL=0.0033mol/L
PH=10 ===> c(OH-)=10^-4 mol/L ===>c(H+)=10^-10mol/L
根據(jù)所得的NaA溶液電荷守恒可知:
c(Na) + c(H+)= c(OH-)+ c(A-)
c(A-)=0.00333mol/L+ 10^-4mol/L -10^-10mol/L=0.00323mol/L
n(A-)=0.00323mol/L×0.030L=9.69×10^-5mol
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也可以用物質(zhì)的量來(lái)計(jì)算:
)用上述NaOH與某一元弱酸HA 20ml反應(yīng),恰好中和時(shí)消耗NaOH溶液10ml,所得溶液PH=10
此時(shí)溶液 n(Na+)= 0.01 mol/L×0.010L=1.0×10^-4 mol
PH=10 ===> c(OH-)=10^-4 mol/L ===>c(H+)=10^-10mol/L
n(OH-)=10^-4 mol/L×0.030L=3.0×10^-6mol
n(H+)=10^-10mol/L ×0.030L=3.0×10^-12mol
根據(jù)所得的NaA溶液電荷守恒可知:
n(Na) + n(H+)= n(OH-)+ n(A-)
n(A-)=1.0×10^-4 mol - 3.0×10^-6mol - 3.0×10^-12mol =9.7×10^-5mol