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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