Give an example based upon particle argument
1 answer:
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The pH of a a 0. 10 m solution of NaCN at 25°C is 11.15
Calculation
The reaction in the solution is given below
→ HCN +
initial 0.1
change ( -x) (+x)
equilibrium ( 0.1 - x ) x
Kb = [HCN] [ ]/[
]
Kb × Ka = Kw = 1.0 ×
Kb = 1.0 × / 4.9 ×
= [HCN] [
]/[
] =
/ ( 0.1 - x )
Kb = 2.04× =
/ ( 0.1 - x )
Since , [NaCN] /Kb > 100 , we can simplify the above equation to
= 2.04× =
/ ( 0.1 )
x = 1.43 × M = [HCN] = [
]
Then pOH = 2.84
pH + pOH = 14
pH = 14- pOH = 14 - 2.84 = 11.15
Therefore , the pH is 11.15
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