Question

In: Chemistry

You wish to titrate a 0.193 M solution of HCN (pKa = 9.24) with a 0.183...

You wish to titrate a 0.193 M solution of HCN (pKa = 9.24) with a 0.183 M solution of NaOH. What is the pH of the HCN solution before the titration?

Expert Solution

HCN before the titration:

HCN <-> H+ +CN-

Ka= [H+][CN-]/[HCN]

in equilibrium:

[H+] = x

[CN-] = x

[HCN] = M-x = 0.193 - x

so

Ka = 10^-pKA = 10^-9.24 = 5.75*10^-10

so, substitute in Ka expression:

5.75*10^-10= x*x/(0.193-x)

solve for x:

x^2 + Ka*x - M*Ka = 0

x = 1.053*10^-5

[H+] = 1.053*10^-5

pH = -log(1.053*10^-5) = 4.977

approx, before any base is added

queen_honey_blossom answered 3 years ago

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