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Sagot :
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
[tex]CN^{-} + H_{2} O[/tex] → HCN + [tex]OH^{-}[/tex]
initial 0.1
change ( -x) (+x)
equilibrium ( 0.1 - x ) x
Kb = [HCN] [[tex]CN^{-}[/tex] ]/[[tex]CN^{-}[/tex] ]
Kb × Ka = Kw = 1.0 × [tex]10^{-14}[/tex]
Kb = 1.0 × [tex]10^{-14}[/tex] / 4.9 × [tex]10^{-10}[/tex] = [HCN] [[tex]CN^{-}[/tex] ]/[[tex]CN^{-}[/tex] ] = [tex]x^{2}[/tex]/ ( 0.1 - x )
Kb = 2.04× [tex]10^{-5}[/tex] = [tex]x^{2}[/tex]/ ( 0.1 - x )
Since , [NaCN] /Kb > 100 , we can simplify the above equation to
= 2.04× [tex]10^{-5}[/tex] = [tex]x^{2}[/tex]/ ( 0.1 )
x = 1.43 × [tex]10^{-3}[/tex] M = [HCN] = [ [tex]OH^{-}[/tex]]
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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