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
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.
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