Okay well I suddenly got confused for a lab I did We took pH meter than we calibrate it. Then we measure the pH of the acetic acid solution, which was used to determine the Ka of the acetic acid using the first method.
We made a 1M solution of NaOH which was used to titrate 50 ml of acetic acid solution, the data which was collected was volume of NaOH versus pH of the solution. The data was ploted and we use second method to find Ka of the solutions.
2.72=-log [H+];
[H+]=0.043M=[Ac-];
To find [HAc] we used the equation
[HAc]ix Va= [NaOH]I x Vb
[HAc]I x 50ml= 0.20M x 25.00ml
[HAc]=0.1 M
Ka for acetic acid (book value) is
Ka= 1.83 x 10^-5
In this method we used Handerson-Hasselbach equation to calculate the Ka of the acetic acid.
pH= pKa + log[Ac-]/[HAc]
BUT NOW I GOT CONFUSED @ FINDING THE pKa is this right? 4.5= pKa
NOW AFTER I FIND THE pKa HOW DO I CALCULATE Ka AGAIN?
pKa = -logKa.
Just enter pKa value, change to - sign, then hit 10x key.
For pKa = 4.75 you will get Ka of 1.778E-5 if you do it right. I just picked a number for you to check your procedure.
how is it that on my calculator I am getting -0.676
pressing -sign then log 4.75 = -0.676
even after I have found the pKa, I don't understand how would I findthe Ka.
Do it this way.
enter 4.75 (that is the pKa.)
change sign to negative to read -4.75. Then PRESS 10X key (not the log key). That would then read 1.75E-5.
sorry. it will read 1.778E-5.
ymzoagc xsawlyue ymguwhpo txszbihe ztjiwupv vfit xpyirz
1 answer