Can someone show me the easisest way to do this problem? How many mL of .120 M of KOH must be added to 60 mL of .20 M HF to produce a solution with a pH of 3.3 ( pka of HF is 3.3).
thanks all !
thanks all !
Can someone show me the easisest way to do this problem? How many mL of .120 M of KOH must be added to 60 mL of .20 M HF to produce a solution with a pH of 3.3 ( pka of HF is 3.3).
thanks all !
This is how i do it.
First of all pH=pka in this question and we know that this is only possible at HALF EQUIVALENCE POINT. So, as usual do MV=MV and whatever volume you get is the EQUIVALENCE POINT and half of that would be HALF EQUIVALENCE POINT, which is when pH=pKa..
In this case, the volume would be equal to 100mL at EQUIVALENCE POINT, so the HALF EQV. POINT would be 50mL and that's where the pH=pKa
Hope that helps.
Hmm question since the NaOH is a strong base and HF a weak acid does it matter in that its going to affect the normal equivalence point?
Let's Review:;
What are the 3 kinds of titrations?
SA and SB pH = 7 at equivalence point
WB and SA pH<7 at equivalence point
WA and SB pH>7 at equivalence point
I hope you get my point. If you see either one of these titrations then you can use MV = MV to get the equivalence volume. If you divide that equivalence volume then you will get the HALF EQUIVALENCE point where the pH of the acid = the pKa of the acid..
Hope that helps.. I am sure you already know this concept but just couldn't think of it..
The normal equivalence pt would be different for all three titrations as I mentioned above..
I already knew that though so the half equivalence point should be different for each one of the scenarios then. And question why do you capitalize Half Equivalence like HALF EQUIVALENCE hahaha