Question

Hello, im having major issues with this problem...I need to find a pH for a 100.0-mL solution of 0.017 M CH3COOH (Ka = 1.8 10-5) is titrated with 0.025 M NaOH.

Now, I've gotten some the answers all ready (right answers have *)....but three of them are wrong, can someone tell me what to do?
1.) initial pH - this is 3.25 *



2.) after addition of 10.0 mL of NaOH- 5.51


3.) after addition of 34.0 mL of NaOH- 4.75*


4.) after addition of 68.0 mL of NaOH-3.29

5.) after addition of 100.0 mL of NaOH-2.32

Answer 1

CH3COOH we will call HAc
100 mL x 0.017 = 1.7 millimoles HAc
10 mL x 0.026 = 0.25 mmoles NaOH.
...........HAc + NaOH ==> NaAx + H2O
initial....1.7....25.......0......0
change....-.25..-0.25......0.25
equil.....1.45.....0........0.25
concn acid = 1.45 mmoles/(100+10)mL=?
concn Ac^- = 0.25/110 = ?

(H^+)(Ac^-)/(HAc) - Ka
Substitute Ac from above and HAc from above and solve for H^+, then convert to pH.

68.0 mL is the equivalence point so pH is determined by hydrolysis of the salt and concn of salt (Ac^-) = 1.7 mmols/168 mL = ?
...........Ac^- + HOH ==> HAc + OH^-
initial...1.7/168..........0.....0
change......-x.............x......x
equil.....you do it........x.......x

Kb = (Kw/Ka) = (x)(x)/(Ac^-)
Substitute Kw, Ka, and Ac^- and solve for x which = OH^-, I would convert to pOH, then to pH.

After 100 mL, you have an excess of OH which is a strong base and pH is determined from the excess of OH added. Don't forget that the total volume will be 200 mL.