Question

In: Chemistry

The pH of 0.30 M solution of HCN is 5.20. Calculate the Ka value for HCN.

Expert Solution

Let a be the dissociation of the weak acid,HA
HA <---> H ⁺ + A^-

initial conc. c 0 0

change -ca +ca +ca

Equb. conc. c(1-a) ca ca

Dissociation constant , K_a = ca x ca / ( c(1-a)

= c a² / (1-a)

In the case of weak acids α is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given c = concentration = 0.30 M

pH = 5.20

- log[H⁺] = 5.20

[H⁺] = 10^-5.20

= 6.31x10^-6 M

[H⁺] = ca = 6.31x10^-6 M

a = (6.31x10^-6 M ) / 0.30

= 2.10x10^-5

So K_a = ca²

= 0.30 x (2.10x10^-5)²

= 1.33x10^-10

queen_honey_blossom answered 1 year ago

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