Answer:
A 0.001 molar solution of weak acid has a pH = 4.0,The pKa is :
B. 5
Explanation:
pH = It is negative logarithm of hydrogen ion concentration.
pH = -log([H+])
Ka = It is the dissociation constant.
pKa = -log(Ka)
The acid dissociates according to the equation given below:
[tex]HA(aq)\rightleftharpoons H^{+}(aq)+A^{-}(aq)[/tex]
[tex]K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}[/tex]
Here , The concentration of A- and H+ will be same
So,
[H+] = [A-].....(1)
Now ,
pH = 4.0
pH = -log([H+])
4.0 = -log([H+])
- 4.0 = log([H+]) take antilog both side::"use scientific calculator"
[tex][H+]=10^{-4}[/tex]
From equation (1), [A-] =
[tex][H+]=[A^{-}]=10^{-4}[/tex]
[HA] = 0.001 M (given)
Insert the value of HA , H+ and A- in Ka
[tex]K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}[/tex]
[tex]Ka=\frac{10^{-4}\times 10^{-4}}{0.001}[/tex]
[tex]Ka=\frac{10^{-8}}{0.001}[/tex]
[tex]Ka=10^{-5}[/tex]
pKa = -log(Ka)
[tex]pKa = -log(10^{-5})[/tex]
pKa = 5
