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Question

Calculate the pH of a 0.10Msolution of NaCN(aq). Ka for HCN is 4.9×1010 at 25oC.

A
11.15
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B
2.85
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C
8.75
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D
7
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Solution

The correct option is A 11.15
CN+H2O=HCN+OH
Initial 0.1
Change (x) (+x)
Equilibrium (0.1x) x

Kb=[HCN][CN]/[CN]

Note Ka×Kb=Kw=1.0×1014
1014/Ka=1.0×1014/4.9×1010

=[HCN][CN]/[CN]=x2(0.1x)
2.04×105=x2(0.1x)

Since [NaCN]/Kb>100, we can simplify the above equation to

2.04×105=x2/0.1
x=1.43×103M=[HCN]=[OH]

Note: pOH=log[OH]

poH=log(1.43×103)=2.84

Note: pH+pOH=14

pH=14pOH
pH=142.84=11.15

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