Question

Determine the pH of a 0.500 M HNO2 solution. Ka of HNO2 is 4.6 * 10-4.

a. 1.82
b. 1.67
c. 2.67
d. 0.30

Answer 1

Answer: pH of solution is = 1.82

Step-by-step explanation:

`HNO_2\rightarrow H^+NO_2^-`

cM 0 0

`c-c\alpha`
`c\alpha`
`c\alpha`

So dissociation constant will be:

`K_a=((c\alpha)^(2))/(c-c\alpha)`

Give c= 0.500 M and
`\alpha` = ?

`K_a=4.6* 10^(-4)`

Putting in the values we get:

`4.6* 10^(-4)=((0.500* \alpha)^2)/((0.500-0.500* \alpha))`

`(\alpha)=0.030`

`[H^+]=c* \alpha`

`[H^+]=0.500* 0.030=0.015`

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

`pH=-log[0.015]=1.82`

Thus pH of a 0.500 M
`HNO_2` solution is 1.82