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

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.

To solve this problem, you need to understand the concept of a weak acid-strong base titration and how to calculate the pH at different stages of the titration.

In this case, you are titrating a weak acid (CH3COOH) with a strong base (NaOH). The reaction that occurs during this titration is:

CH3COOH + NaOH -> CH3COONa + H2O

To find the pH at different stages of the titration, you need to calculate the moles of acid and base present, and then use the Henderson-Hasselbalch equation to calculate the pH.

1.) To find the initial pH, you need to calculate the moles of CH3COOH in 100.0 mL of the 0.017 M solution. The number of moles can be calculated using the formula:

moles = concentration x volume

moles of CH3COOH = 0.017 M x 0.100 L = 0.0017 moles

Next, you need to use the Henderson-Hasselbalch equation, which is given by:

pH = pKa + log([A-]/[HA])

In this case, [A-] represents the concentration of the conjugate base (CH3COO-), and [HA] represents the concentration of the acid (CH3COOH). The pKa value is given as 1.8 x 10^-5.

Since the initial solution only contains CH3COOH, the concentration of the conjugate base is 0 and the concentration of the acid is the same as the initial concentration (0.017 M). Plugging these values into the Henderson-Hasselbalch equation, we get:

pH = 4.74 + log(0/0.017)

Since log(0) is not defined, we assume that the concentration of CH3COOH is much greater than the concentration of CH3COO-. Therefore, the pH is simply equal to the pKa value, which is 4.74.

So, the correct answer for the initial pH is 4.74, not 3.25 as you mentioned.

To find the pH at other stages of the titration, you need to consider the volume of NaOH added and the stoichiometry of the reaction. You can calculate the moles of NaOH added and then determine how many moles of CH3COOH and CH3COO- are present.

Modify the previous steps accordingly for each stage and calculate the pH using the Henderson-Hasselbalch equation. Make sure to take into account the volumes of both CH3COOH and NaOH at each stage.

I hope this explanation helps you understand the correct approach to solving this problem!