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meenugouri

45 ml of 0.0325 M HCN solution is titrated with 0.0200 M NaOH solution.( Ka of HCN =6.2 x 10 power - 10 )

1) calculate volume of NaOH used in titration in order to reach equivalence point ?

2) What is the pH of the solution at equivalent point ?

3) What is molar concentration of CN- at equivalent point ?

i.rahul.mehra

millieq = N * V

N^₁ V₁= N₂V₂

V ₂= 0.0325 * 45 / 0.0200

= 72.8 ML

krishna.gopal

At equivalent point concentration of NaCN (c) = is 0.0325*45/(45+72.8)

If h is degree of hydrolysis then for recation
CN- + H2O --> HCN + OH-
equllibrium constant is Kh= Kw/Ka
Conc of [OH-]=ch
Conc of [HCN] = ch
conc of [CN-] = c(1-h)
if h<<1 then
h = sqrt(Kw/(Ka*c))
So [OH-] = sqrt(cKw/Ka)
use this to find pH
Conc. of [CN-] = c(1-h)

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