Question

常温下,将0.06mol.L的硝酸溶液和未知浓度的KOH溶液以1:5的体积比混合,所得溶液的PH=12.用上述KOH溶液

Answer 1

HNO3 +KOH =KNO3 +H2O

1 1

V*0.06 5V*X

(5V*X-0.06V)=10^-2*(V+5V)

X=0.024mol/L

HA +KOH =KA +H2O

1 1

C*0.12 0.05*0.024

C=0.01mol

HA ==可逆= H+ + A-

0.01

0.01-0.001 0.001 0.001

K=0.001*0.001/(0.01-0.001)

= 1.11*10^-4

答 :略