Chemistry homework help?
2016-10-24 4:21 pm
Calculate the pH of a buffer solution that is 0.250 M in HCN and 0.170 M in KCN. For HCN, Ka=4.9×10−10 (pKa=9.31). Use both the equilibrium approach and the Henderson-Hasselbalch approach.
回答 (1)
2016-10-24 5:23 pm
Equilibrium: HCN <--> H+ + CN-
Ka = [H+][CN-]/[HCN] = 4.9X10^-10
[H+] (0.170) / (0.250) = 4.9X10^-10
[H+] = 7.2X10^-10
pH = 9.14
Using H-H equation:
pH = pKa + log [CN-]/[HCN]
pH = 9.31 + log (0.170/0.250) = 9.14
Ka = [H+][CN-]/[HCN] = 4.9X10^-10
[H+] (0.170) / (0.250) = 4.9X10^-10
[H+] = 7.2X10^-10
pH = 9.14
Using H-H equation:
pH = pKa + log [CN-]/[HCN]
pH = 9.31 + log (0.170/0.250) = 9.14
收錄日期: 2021-04-21 23:45:36
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