Answer:
The answer is
[tex]Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3} [/tex]
Explanation:
The Ka of an acid when given the pH and concentration can be found by
[tex]pH = - \frac{1}{2} log(Ka) - \frac{1}{2} log(c) [/tex]
where
c is the concentration of the acid
From the question
pH = 5.82
c = 0.010 M
Substitute the values into the above formula and solve for Ka
We have
[tex]5.82 = - \frac{1}{2} log(Ka) - \frac{1}{2} log(0.010) [/tex]
[tex] - \frac{1}{2} log(Ka) = 5.82 + 1[/tex]
[tex] - \frac{1}{2} log(Ka) = 6.82[/tex]
Multiply through by - 2
[tex] log(Ka) = - 13.64[/tex]
Find antilog of both sides
We have the final answer as
[tex]Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3} [/tex]
Hope this helps you
