Posted by **Sarah** on Thursday, September 29, 2011 at 4:21pm.

Suppose that 30.0 mL of 0.20 M C6H5COO(aq) is titrated with 0.30 M KOH(aq).

a) What is the initial pH of the 0.20 M C6H5COOH(aq)?

b) What is the pH after the addition of 15.0 mL of 0.30 M KOH(aq)?

c) What volume of 0.30 M KOH(aq) is required to reach halfway to the stoichiometric point?

d) Calculate the pH at the halfway point.

e) What volume of 0.30 M KOH(aq) is required to reach the stoichiometric point?

f) Calculate the pH at the stoichiometric point.

Please show work as well as correct answers. Thanks!

- CHEMISTRY -
**DrBob222**, Thursday, September 29, 2011 at 4:40pm
Most of these titration problems are worked the same way. The secret is to know where you are on the titration curve.

a. You have a solution of C6H5COOH. Use Ka, set up an ICE chart, substitute, solve for H^+ and convert to pH.

b. initial moles acid = M x L

Subtract moles KOH added = M x L

Convert moles of each component to M and substitute into the Ka expression OR substitute into the Henderson-Hasselbalch equation.

c. Same thing for half way point except you need to determine where the half wa point it. Calculate how much KOH it takes to move to the equivalence point, then take half that number of mL and work it the same way as part b.

d. You've worked this part if you followed the instructions in part c. The stoichiometric point and the equivalence point are the same thing.

Post your work if you get stuck.

- CHEMISTRY -
**Sarah**, Thursday, September 29, 2011 at 8:38pm
For A) I got the correct answer from my ICE table (2.44)

For B) I got 0.006 moles of C6H5COOH initially and 0.0045 moles of KOH was added. The excess C6H5COOH is 0.006 - 0.0045 = 0.0015 moles. I then converted the 0.0015 moles to M (mol/L) by multiplying by 1000/30. This is the concentration of C6H5COOH multiply by the Ka value and sqrt to find the [H3O+] concentration. My answer is 2.74 but it's incorrect.

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