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A 75.0-mL sample of 0.0500 M HCN (Ka = 6.2 × 10–10) is titrated with 0.220...

A 75.0-mL sample of 0.0500 M HCN (Ka = 6.2 × 10–10) is titrated with 0.220 M NaOH. What is the [H+] in the solution after 3.0 mL of 0.220 M NaOH have been added?

a. 3.4 × 10–6 M

b. '4.7 M

c. 2.9 × 10–9 M

d.1.0 × 10–7 M

e. None of these choices are correct.

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Answer #1

millimoles of HCN = 75 x 0.05 = 3.75

millimoles of NaOH added =3.0 x 0.220 = 0.66

3.75 - 0.66 = 3.09 millimoles HCN left

0.66 millimoles salt formed

[acid] = 3.09 / 80 = 0.0386 M

[salt] = 0.66 / 80 = 0.00825 M

pH = pKa + log [salt] / [acid]

pKa = - log Ka = - log [6.2 x 10-10]

pKa = 9.21

pH = 9.21 + log [0.00825] / [0.0386]

pH = 8.54

[H+] = 10-pH = 10-8.54 = 2.9 x 10-9 M

answer = option c

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