A 0.0170 M solution of an acid has a pH of 2.5. What is the pKa?
i did 10^-2.5=3.16e-3
Then 0.0170 - 3.16e-3=0.01384
ka= (3.16e-3)^2 / 0.01384 = 7.21e-4
pKa = -log(7.21e-4) = 3.14
The pKa is 3.14
Would you agree?
Very good work.
Thank you!
please can I get a more detailed solution
Yes, I agree with your calculation. To find the pKa, you started by calculating the concentration of the acid using the pH value. You used the formula 10^-pH to find the concentration, which gave you 3.16e-3 M.
Next, you subtracted this concentration from the initial concentration of the acid solution (0.0170 M) to determine the concentration of the conjugate base. This gave you 0.01384 M.
Then, you used the formula for the acid dissociation constant (Ka), which is Ka = [conjugate base]^2 / [acid]. You plugged in the values and calculated Ka to be 7.21e-4.
Finally, you calculated the pKa by taking the negative logarithm (base 10) of Ka. This gave you a pKa value of 3.14.
Therefore, based on your calculations, the pKa of the acid is indeed 3.14.