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.
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
1 answer
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