Hello, im having major issues with this problem...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

I'm suppose to be finding out the pH of each solution...sry

Sure, I can help you with that. In order to determine the pH at different points during the titration process, you need to consider the stoichiometry and equilibrium of the reaction between CH3COOH (acetic acid) and NaOH (sodium hydroxide).

Here's how you can calculate the pH at each step:

1.) Initial pH (before any NaOH is added):
To find the initial pH, you can use the equation for the dissociation of acetic acid:
CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+ (aq)
The initial concentration of CH3COOH is given as 0.017 M. Since acetic acid is a weak acid, we can assume that its concentration remains constant during the titration. Thus, the concentration of H3O+ ions can be calculated using the acid dissociation constant (Ka).
Ka = [CH3COO-][H3O+] / [CH3COOH]
Assuming the concentration of H3O+ at the start is negligible compared to the concentration of CH3COOH, we can approximate it as zero.
Ka = [CH3COO-][H3O+] / [CH3COOH] ≈ [CH3COO-][H3O+] / [CH3COOH]
Since the initial concentration of CH3COOH is given as 0.017 M, we can rearrange the equation to solve for [H3O+]:
[H3O+] = (Ka * [CH3COOH])^0.5 = (1.8 x 10^-5 * 0.017)^0.5 ≈ 3.25 x 10^-3 M
Taking the negative logarithm of the [H3O+] concentration, we get the initial pH:
pH = -log[H3O+] = -log(3.25 x 10^-3) ≈ 2.49

However, you mentioned that you obtained an answer of 3.25 for the initial pH. Double-check that you calculated the concentration of H3O+ correctly and took its negative logarithm to get the pH.

2.) After addition of 10.0 mL of NaOH:
At this point, 10.0 mL of a 0.025 M NaOH solution is added to the acetic acid solution. This will react with the acetic acid based on a 1:1 stoichiometry:
CH3COOH (aq) + NaOH (aq) → CH3COONa (aq) + H2O (l)
The moles of CH3COOH that have reacted can be calculated by multiplying the volume of NaOH solution added (in L) by its molarity:
moles of CH3COOH reacted = Volume of NaOH added (in L) * Concentration NaOH
= 10.0 mL / 1000 mL/L * 0.025 mol/mL = 2.5 x 10^-4 mol
Since the initial moles of CH3COOH were based on a 100.0 mL volume, you need to subtract the moles that have reacted:
moles of CH3COOH remaining = Initial moles of CH3COOH - moles of CH3COOH reacted
= 0.100 L * 0.017 mol/L - 2.5 x 10^-4 mol
To find the concentration of acetic acid after the reaction, divide the moles by the final volume (100.0 mL + 10.0 mL):
Concentration of CH3COOH = moles of CH3COOH remaining / Volume (L)
= (0.017 mol - 2.5 x 10^-4 mol) / (0.100 L + 0.010 L)
Now, you can use the equation from the first step to determine the [H3O+] concentration and thus the pH.

Repeat the same process for the subsequent steps by considering the volume of NaOH added and adjusting the concentration of CH3COOH accordingly. Calculate the moles of CH3COOH reacted and the remaining concentration of CH3COOH, and then use the equation from the first step to find the [H3O+] concentration and pH.

Let me know if you need further assistance with the calculations.