Question

2.计算0.01 molL-1HF溶液的[H+]和解离度已知HF

Answer 1

问：2.计算0.01 molL-1HF溶液的[H+]和解离度已知HF

答：HF在水溶液中的解离平衡式为 HF H+ + F-HF的解离常数为Ka =6.6×10-4因为 C/Ka=0.1/(6.6×10-4)>500 (1分)所以[H+]=(C×Ka)1/2=(0.10×6.6×10-4)1/2=0.0081mol/L

答：PH =-lg[H+]=2.1 (2分)Ka/C1/2=([H+]/C)×100%=8.1%

答：PH =-lg[H+]=2.1 Ka/C1/2=([H+]/C)×100%=8.1%