hi im having problems figuring this out.

consider the titration of 100 ml of 0.200 acetic acid by 0.100 koh. Calculate the ph if 250 ml of koh is added ?

I tried to find the limiting reagent and go from there but it was a no go

then i used the kb for find the poh and then the ph..i got 8.27

answer is 12.15

what is the correct method ?

You started correctly. I don't know where you went wrong.

mols acetic acid initially = 0.1 x 0.2 = 0.02.
mols KOH at 250 = 0.250 x 0.1 - 0.025
So all of the acetic acid is used; an excess of 0.005 mols KOH remains. The volume (which probably is what you failed to take into account) is 100 + 250 = 350 mL.
So (OH^-) = 0.005/0.350 = ??
pOH and pH from there. I get 12.15, too.

thanks

i found the lr

then plugged did some extra stuff that i should have

thanks again Jim

how bout if 200ml of koh is added ?

how bout if 200ml of koh is added ?

the answer is 8.78

im getting 12.82

At 200 mL, you are at the equivalence point.

acetic acid KOH ==> Kacetate + H2O

acetic acid initially = 0.02 moles
KOH added 0.2 L x 0.1 M = 0.02 moles.
So what do we have in solution. It is Kacaetate (potassium acetate) in 300 mL water. What determines the pH of salts in water. Its the hydrolysis, of course.
So, acetate ion, which is C2H3O2^- hydrolyzes as follows:
C2H3O2^- + HOH --> HC2H3O2 + OH^-
Then Kb = Kw/Ka = (HAc)(OH^-)/(C2H3O2^-)
Set up an ICE chart and calculate OH^- and from there pH.
I get 8.79 using 1.75 x 10^-5 for Ka.

thanks.

Damn your smart dr.bob

To calculate the pH of a solution after a titration, you need to know the balanced chemical equation for the reaction and use the concept of stoichiometry. Here's the correct method step by step:

Step 1: Write the balanced chemical equation for the reaction between acetic acid (CH3COOH) and KOH. In this case, it is:

CH3COOH + KOH -> CH3COOK + H2O

Step 2: Determine the moles of acetic acid in the 100 mL solution. To do this, you can use the molarity equation:

moles of solute (acetic acid) = volume (in liters) x concentration

Since the volume is given in milliliters, you need to convert it to liters:

Volume (in liters) = 100 mL / 1000 mL/L = 0.1 L

Using the concentration provided (0.200 M), you can calculate the moles:

moles of acetic acid = 0.1 L x 0.200 M = 0.02 moles

Step 3: Determine the moles of KOH added. Again, you can use the molarity equation:

moles of solute (KOH) = volume (in liters) x concentration

The volume is given as 250 mL, so you'll need to convert it to liters:

Volume (in liters) = 250 mL / 1000 mL/L = 0.25 L

Using the concentration provided (0.100 M), you can calculate the moles:

moles of KOH = 0.25 L x 0.100 M = 0.025 moles

Step 4: Determine the limiting reagent. Compare the mole ratios from the balanced equation. You'll see that the ratio of moles of acetic acid to KOH is 1:1.

Since the moles of acetic acid (0.02 moles) are less than the moles of KOH (0.025 moles), the limiting reagent is acetic acid. This means that all the acetic acid will react, and there will be some excess KOH remaining after the reaction.

Step 5: Calculate the remaining moles of KOH. Since the limiting reagent is acetic acid, the KOH is in excess. Subtract the moles of acetic acid reacted from the total moles of KOH added:

moles of KOH remaining = moles of KOH added - moles of acetic acid reacted
= 0.025 moles - 0.02 moles
= 0.005 moles

Step 6: Calculate the concentration of KOH remaining. Divide the moles of KOH remaining by the volume of the solution (350 mL = 0.35 L in total after the addition of KOH):

concentration of KOH remaining = moles of KOH remaining / volume of solution
= 0.005 moles / 0.35 L
= 0.014 M

Step 7: Find the pOH of the solution. From the balanced chemical equation, you can see that KOH is a strong base, so you can use the concentration of KOH remaining to find the pOH:

pOH = -log10(concentration of KOH remaining)
= -log10(0.014)
= 1.85

Step 8: Find the pH of the solution. Since pH + pOH = 14 (at 25°C), you can calculate the pH:

pH = 14 - pOH
= 14 - 1.85
= 12.15

Therefore, the correct pH after the addition of 250 mL of KOH is 12.15.