pls confirm:
pH=-log(0.0069) = 2.158
pH=14-2.156 = 11.8
Lactic acid is a weak acid with the formula , HCH3H5O3, the Ka for lactic acid is 1.38 x 10-4.
In aqueous solution, lactic acid partially dissociates according to the following reaction:
HCH3H5O3 ⇔ CH3H5O3- + H+
Use the Ka equation to calculate the pH of the lactic acid solution described below:
Volume: 125 mL
Concentration: 0.3494 M
Since this is a weak acid, you can assume the amount of acid dissociated is >> 5% of the total amount of acid present.
5 answers
No and you should know by intuition it can't be correct. A pH of 11.8 is quite basic and you know an acid solution MUST be < 7.0.
You answer of 2.15 is OK so why did you go through the last step (which calculate pOH)? They asked for pH and not pOH. In the future you should show your calculations. We can check calculations much faster than sitting down with pen and paper and calculating the answer so we can compare it with yours.
You answer of 2.15 is OK so why did you go through the last step (which calculate pOH)? They asked for pH and not pOH. In the future you should show your calculations. We can check calculations much faster than sitting down with pen and paper and calculating the answer so we can compare it with yours.
ok so it will only be subtracted from 14 if we need pOH? otherwise we should just leave it as 2.15 in this case. thanks for your help
Yes. pH = -log(H^+) give you the pH and that's the question in the problem.
Then using pH + pOH = pKw = 14 and solving for pOH is another step for another problem (which in this case they didn't ask).
Then using pH + pOH = pKw = 14 and solving for pOH is another step for another problem (which in this case they didn't ask).
What information is needed in order to calculate Ka?
1 answer