Question

What is the pH of 0.20 M nitrous acid at equilibrium (Ka=4.6 x 10 -4)

Answer 1

Answer is A on Plato

Answer 2

Answer : The pH of the solution is, 2.03

Solution : Given,

Concentration (c) = 0.20 M

Acid dissociation constant =
`k_a=4.6* 10^(-4)`

The equilibrium reaction for dissociation of
`HNO_2` (weak acid) is,

`HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+`

initially conc. c 0 0

At eqm.
`c(1-\alpha)`
`c\alpha`
`c\alpha`

First we have to calculate the concentration of value of dissociation constant
`(\alpha}`.

Formula used :

`k_a=((c\alpha)(c\alpha))/(c(1-\alpha))`

Now put all the given values in this formula ,we get the value of dissociation constant
`(\alpha}`.

`4.6* 10^(-4)=((0.2\alpha)(0.2\alpha))/(0.2(1-\alpha))`

By solving the terms, we get

`\alpha=0.0468`

No we have to calculate the concentration of hydronium ion or hydrogen ion.

`[H^+]=c\alpha=0.2* 0.0468=9.36* 10^(-3)M`

Now we have to calculate the pH.

`pH=-\log [H^+]`

`pH=-\log (9.36* 10^(-3))`

`pH=2.03`

Therefore, the pH of the solution is, 2.03