Question

What concentration of NaCN must be added to a 0.5 M HCN solution to produce a buffer solution with pH 7.0? K_a = 6.2 x 10^–10 for HCN

a) 3.3 M

b) 0.49 M

c) 6.9 x 10^-5 M

d) 0.0031 M

e) 0.22 M

Answer 1

To solve this we need to use Henderson-Hasselbalch equation which is given by,

pH = pKa + log([base]/[acid])

Since Ka = 6.2 * 10-10

Therrefore pKa = -log(Ka) = -log(6.2 * 10-10) = 9.2076

Therefore

7.0 = 9.2076 + log([base]/0.5)

7.0 - 9.2076 = log([base]/0.5)

-2.2076 = log([base]/0.5)

Take antilogarithm on both side

10-2.2076 = ([base]/0.5)

0.0062 = ([base]/0.5)

Multiply both side by 0.5

0.0062 * 0.5 = [base]

0.0031 M = [base]

Therefore concentration of NaCN added to make buffer solution is 0.0031 M.

If you find any mistake, please mention in the comment box. Thanks.