Description
An aqueous solution of a monoprotic acid HA (Ka = 1.45×10-1) has a pH of 4.75.
What is the molarity of the solution? (i.e. [HA]+[A-].)
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Explanation & Answer
pH = 4.75
[H+] =10 ^ -4.75 = 1.78e-5
HA | H+ | A- | |
Initial | M | 0 | 0 |
Change | -x | +x | +x |
Equilibrium | M -1.78e-5 | 1.78e-5 | 1.78e-5 |
1.45e-1 M -2.581e-6 = 3.1684e-10
M = 1.78e-5M
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