Question

Henderson-Hasselbalch: pH=pka + Log( [base]/(acid] ) 1. The value of Ka of nitrous acid (HNO2) is 4.6 x 104 M. Calculate the pH and the concentration of [H3O+] in a 0.02M aqueous solution of HNO2(aq). 2. Label conjugate acid-base pairs: HNO2(aq) + H2O → H30* + NO2 (aq) 3. What will happen to the reaction equilibrium if we increase the pressure in the reaction vessel? H2(g) + 12(e) → 2 HI(g)

Need help on questions 1-3

Answer 1

1. given Ka =4.6*10^(-4)M

[HNO2] = 0.02M

now for acid

HNO2 + H2O $\rightleftharpoons$ H3O+. + NO2-

let [H3O+] = [NO2-] =x

Ka is written as

Ka = [H3O+] [NO2-]/[HNO2] = 4.6*10^(-4) M. (given)

thus x. x /0.02 = 4.6*10^(-4)

x.x = 9.2* 10^(-6)

thus x= [H3O+] = 3.03 * 10^(-3)

now pH = - log [H3O+] = - log(3.03*10^(-3)) = 2.52

2. HNO2 +H2O  $\rightleftharpoons$ H3O+ + NO2-

here HNO2 is acid as it is lossing H+ ion in water and thus its conjugate base is NO2-.

Here H2O is acting as a base, because it is accepting a proton. thus H3O+ is its conjugate acid.

3. H2 +I2 $\rightleftharpoons$ 2HI

As pressure increases, volume decreases. then no. of moles per unit volume increases. To overcome this effect, equilibrium will shift in a direction of decreasing no. of moles.

but here no. of moles are same in both side. So here no effect of prrssure change.