Question

What is the pH of an aqueous solution that contains 0.282 M NaCN and 0.724 M HCN?

            K_a(HCN) = 4.9 x 10^-10

	a.	pH = 9.72
	b.	a) pH = 8.90
	c.	pH = 13.23
	d.	pH = 20.49
	e.	pH = 9.31

Answer 1

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