All of these look like homework dump to me. You show no idea of your own, how you think you could solve the problem yourself or even how you might procede. I've worked all of the problems anyway. In the future you should give me some indication of the work you have done and exactly what you don't understand about the problem.Remember that this is a homework HELP with the emphasis on HELP part. It is not my intention to take this chemistry course again. I already know how to work these problems. So I want to HELP but I don't want to DO them all.
I will call HC2H3O2 HA to save typing. NaA will be sodium acetate.
millimoles HAc = 25 mL x 0.1 M = 2.5 mmols
mmols NaOH added....25 mL x 0.1 M = 2.5 mmols
......................HAc + NaOH --> NaAc + H2O
I..................... 2.5..........0............0............0
add...............................2.5..............................
C.....................-2.5......-2.5...........+2.5................
E......................0...........0...............2.5.............
So after 25.00 are added you have aolution of Ac^- (from NaAc) and the pH depends upon the hydrolysis of the Ac^- as follows:
(Ac^-) = millimoles/mL = (2.5 mmols/50 mL) = 0.05M
.........................Ac^- + HOH ==> HAc + OH^-
I.......................0.05........................0...........0
C.......................-x............................x............x
E......................0.05-x........................x...........x
Kb for Ac^- = Kw/Ka for HAc = (x)(x)/(Ac^-)
Solve for OH, convert to pOH and then to pH as I've shown in one or two of the problems below.
Or there is a shortcut way of doint it directly
(H^+) = sqrt (KwKa/C) where C = (Ac^-) = 0.05 M
A 25.00mL sample of 0.100M HC2H3O2 is titrated with 0.100M NaOH. What is the pH after 25.00mL of NaOH have been added? Ka HC2H3O2 = 1.8x10-5.
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Unlike DrBOB222 I WILL SPOON FEED YOU ANSWER PLEASE LET ME KNOW ANYTHING WRONG I AM WEEK AT ENGLISH SO PLEASE LET ME KNOW...NO REQUIREMENT CALCULUS BCDEF.
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