In: Chemistry
What is the pH of an aqueous solution that contains 0.282 M NaCN and 0.724 M HCN?
Ka(HCN) = 4.9 x 10-10
a. |
pH = 9.72 |
|
b. |
a) pH = 8.90 |
|
c. |
pH = 13.23 |
|
d. |
pH = 20.49 |
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e. |
pH = 9.31 |
HCN is a weak acid,and NaCN is a salt with strong base,so it act as a buffer.
According to Henderson equation pH of buffer is
pH=pKa+log{[Salt]/[Acid]}=pKa+log{[NaCN]/[HCN]}
pKa=-logKa
Ka=4.9X10^-10
pKa=-log[4.9X10^-10]=-[-9.309]=9.309=9.31( after rounding off)
[NaCN]=concentration of salt,that is 0.282 M
[HCN]=concentration of acid,that is 0.724 M
pH=pKa+log{[NaCN]/[HCN]}
=9.31+log{[0.282 M]/[0.724 M]}
=9.31+log[0.3895]
=[9.31-0.409]=8.901=8.90(after rounding off)
Answer :pH=8.90