An aqueous solution of a monoprotic acid is prepared. This acid is known to have a pKa of 8.6. If the acid is 37.5% ionized when in solution, what is the pH of the solution?
Not really sure where to start with this question? Thanks in advance
pKa = -log Ka. Solve for Ka.
Then..........HA ==> H^+ + A^-
I.............1.......0.....0
C...........-.375...0.375..0.375
E...........0.625...0.375..0.375
Substitute the E line into the Henderson-Hasselback equation and solve for pH.
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To determine the pH of the solution, we need to know the concentration of the acid in solution and the degree of ionization.
Given that the acid is monoprotic, it donates one proton (H+) when it ionizes. Therefore, the concentration of the acid, [A], is equal to the concentration of H+ ions in solution.
Let's assume we have a 100 mL solution of the acid. Based on the statement that the acid is 37.5% ionized, we can determine that 37.5 mL of the acid has ionized, bringing 37.5 mL of H+ ions into solution.
Now, we need to determine the concentration of the acid in solution. Since we started with 100 mL of solution, and 37.5 mL ionized, the remaining 62.5 mL is not ionized. Therefore, [A] = (37.5 mL / 100 mL) * 100% = 37.5% = 0.375.
The remaining step is to determine the pH of the solution using the equation for pKa:
pH = pKa + log([A]/[HA])
Given that the pKa is 8.6, and that [A] = 0.375, the equation becomes:
pH = 8.6 + log(0.375/[HA])
However, we are given that [A] = [HA] since the acid is 37.5% ionized. Therefore, the equation becomes:
pH = 8.6 + log(0.375/0.375) = 8.6 + log(1) = 8.6
Hence, the pH of the solution is 8.6.