Calculate the pH of a 0.036 M nitrous acid (HNO2) solution (Ka HNO2 = 4.5 x 10^-4)?

Use an ICE table. For a monoprotic weak acid this has a standard result

Ka = x^2/([I]-x)

you know Ka, initial concentration of the acid [I]= 0.036, so solve for x = [H+]

from x calculate pH

-0.60

Ka=([H+] [A-]) / ([HA]) HA:your acid

Ka=([H+] [NO2-]) / ([HNO2]) now put your known parameters in equation. you shall have:

Ka=4.5 * 10e-4 [HNO2]=0.036

NOTE!:because nitrous acid s dissociaton coff. is 1 thus it completely dissolves in water and concentration of NO2- will be the same with concentration of HNO2. you now have: [NO2-]=0.036

now you can rewrite the equation:

4.5 * 10e-4 = ([H+] * 0.036) / (0.036)

[H+] = 4.5 * 10e-4

pH= - log (4.5 * 10e-4) =3.346

HNO2----> H{+} + NO2{-}

C-aC aC aC

Ka=[H+][NO2-]/[HNO2]--->Ka=a^2 x C/ 1-a

because Ka/C larger of 0,01 you solve quadratic equation

4,5x10^-4=a^2x0,036/1-a--->0,0125=a^2/1-a--->

(solving the quadratic equation you find) a=0,106

[H+]=aC, a=0,106 ,C=0,036M --->[H+]=3,82x10^-3

pH=-log[H+]=-log[3,82x10^-3]=2,42

pKa= -log(4.5e-4)= 3.35

pH= 0.5(pKa- log[0.036])

pH= 0.5(3.35 + 1.44)

pH= 4.07

1.44?

idk