What is the pH of a .20M HA solution (Ka=1.0x10^-6) that contains a 0.40M NaA?
I found that it came to 6.30 for the pH. So I first balanced the equation set up an "ICE" box and then solved for x using the Ka equation...Ka=([H+][A-])/([HA]), I neglected the +x and -X and plugged in to solve. Then I got the pH by doing the negative log of the x. Hope this helped :)
I agree with 6.30 but I used the Henderson-Hasselbalch equation. It's easier and you don't need an ICE chart.
To find the pH of the solution, we need to consider the dissociation of the weak acid, HA.
The dissociation reaction is written as follows:
HA ⇌ H+ + A-
From the given information, we know that the concentration of HA is 0.20 M, and that it dissociates to form H+ and A-. Additionally, we have the equilibrium constant Ka, which is equal to [H+][A-]/[HA]. For a weak acid, we can assume that the concentration of H+ released from the dissociation of HA is approximately equal to the concentration of A-.
Using the equilibrium constant expression, we can write:
Ka = [H+][A-]/[HA]
Since [H+] ≈ [A-], we can substitute [HA] = 0.20 M - [H+] into the equation:
Ka = [H+][A-]/(0.20 M - [H+])
Rearranging the equation gives:
[H+]^2 - Ka[H+] + Ka(0.20 M) = 0
To solve this quadratic equation, we can use the quadratic formula:
[H+] = (-b ± √(b^2 - 4ac))/(2a)
Applying this formula, we have:
[H+] = (Ka ± √(Ka^2 - 4(1)(Ka)(0.20 M)))/(2(1))
Simplifying and evaluating the expression, we find the concentration of H+ in the solution. Finally, we can determine the pH by using the equation: pH = -log[H+].