Asolution contains a weak monoprotic acid HA and
its sodium salt NaA both at 0.1 M concentration.
Show that [OH�] Kw/Ka.
Start with an equation
HA <-> H+ + A-
at start
0.1.....0....0
at equilibrium
0.1-x...x....x+0.1
Ka=[H+][A-]/HA
=x.(0.1+x)/(0.1-x)
if x is small with respect 0.1 then this expression becomes
Ka=x.(0.1)/(0.1)
or Ka=x
Kw=[H+](OH-]
=x[OH-]
so
[OH-]=Kw/x
as x=Ka then
[OH-]=Kw/Ka
To show that [OH-] = Kw/Ka, we need to calculate the concentration of hydroxide ions ([OH-]) in the solution and compare it to the expression Kw/Ka.
1. Start by writing down the balanced equation for the dissociation of the weak monoprotic acid HA:
HA ⇌ H+ + A-
2. The acid dissociation constant (Ka) is given as the ratio of the concentrations of the products (H+ and A-) to the concentration of the undissociated acid (HA):
Ka = [H+][A-] / [HA]
3. Since the concentrations of HA and its sodium salt NaA are both 0.1 M, we can substitute these values into the Ka expression:
Ka = [H+][A-] / 0.1
4. The product of hydroxide ion concentration ([OH-]) and hydrogen ion concentration ([H+]) in water is given by the ion product of water (Kw), which is equal to 1.0 × 10^-14 at 25°C. Therefore, [H+][OH-] = Kw.
5. Rewrite the equation Kw = [H+][OH-] and solve for [OH-]:
[OH-] = Kw / [H+]
6. In the case of a weak monoprotic acid, we can assume that [H+] = [A-], since they are formed in a 1:1 ratio during the dissociation of the acid.
7. Substitute [A-] for [H+] in the expression for [OH-]:
[OH-] = Kw / [A-]
8. Since [A-] = 0.1 M, we can substitute this value into the expression:
[OH-] = Kw / 0.1
9. By dividing both sides of the equation by 0.1, we obtain:
[OH-] = 10 × (Kw / 0.1)
10. Finally, recall that Kw/Ka = 1.0 × 10^-14 / Ka. So, substitute this expression into the equation:
[OH-] = 10 × (Kw/Ka)
Therefore, we have shown that [OH-] = Kw/Ka.