Question

# A 75.0-mL sample of 0.0500 M HCN (Ka = 6.2 ×10-10) is titrated with 0.273 M...

A 75.0-mL sample of 0.0500 M HCN (Ka = 6.2 ×10-10) is titrated with 0.273 M NaOH. What is [H+] in the solution after 3.0 mL of 0.273 M NaOH has been added?

A.

4.5 × 10-6M

B.

1.0 × 10-7M

C.

3.6 M

D.

2.2 × 10-9M

E.

none of these

we know that

moles = molarity x volume (L)

so

moles of HCN = 0.05 x 75 x 10-3 = 3.75 x 10-3

moles of NaOH added = 0.273 x 3 x 10-3 = 0.819 x 10-3

now

the reaction is

HCN + NaOH ----> NaCN + H20

we can see that

moles of HCN reacted = moles of NaOH added = 0.819 x 10-3

moles of NaCN formed = moles of NaoH added = 0.819 x 10-3

now

finally

moles of HCN = 3.75 x 10-3 - 0.819 x 10-3 = 2.931 x 10-3

moles of NaCN = 0.819 x 10-3

now

HCN and NaCN are a buffer combination

for buffers

pH = pKa + log [ salt / acid ]

also

pKa = -log Ka

so

pH = -log Ka + log [ NaCN / HCN]

pH = -log 6.2 x 10-10 + log [ 0.819 x 10-3 / 2.931 x 10-3 ]

pH = 8.654

now

pH = -log [H+]

so

-log [H+] = 8.654

[H+] = 2.21 x 10-9

so

the answer is D) 2.2 x 10-9 M

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