CHEMISTRY

posted by .

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 -

    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 -

    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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chemistry

    A 0.25 mol/L solution of benzoic acid, KC7H5O2 and antiseptic also used as a food preservative, has a pH of 2.40. Calculate the Ka of benzoic acid at SATP My work: C6H5COOH --> C6H5COO^- + H^+ Ka = (C6H5COO-)(H+)/ C6H5COOH I know …
  2. Chemistry, acids and bases

    (a) Find pH and pOH of 1.0 M solution of sodium benzoate, NaC6H5COO. (Ka of benzoic acid is 6.2x10^-5) (b) Calculate pH after 0.205 moles per liter of HCl was added (assume volume does not change (c) Calculate pH after 1.0 moles per …
  3. Chemistry

    Suppose that 30.0 mL of 0.20 M C6H5COOH(aq) is titrated with 0.30 M KOH(aq). Refer to table 1 and table 2. (a) What is the initial pH of the 0.20 M C6H5COOH(aq)?
  4. Chemistry

    Suppose that 30.0 mL of 0.20 M C6H5COOH(aq) is titrated with 0.30 M KOH(aq). (a) What is the initial pH of the 0.20 M C6H5COOH(aq)?
  5. chemistry

    1) a 100ml smaple of 0.18M HCLO4 is titrated with 0.27M LiOH. determine the ph of the soution after the addition of 66.67 ml of LiOH (this is at equivalence point) 2)a 100 ml sample of .20M HF is titrated with .10M KOH. determine th …
  6. chemistry

    0.5 L of a 0.30 M HCl solution is titrated with a solution of 0.6 M KOH. a)What is the pH before addition of KOH?
  7. Chemistry

    C6H5COOH(aq) + NO2-(aq) <=> C6H5COO-(aq) + HNO2(aq) A table of ionizations is given to calculate the equilibrium constant for this overall reaction.. but when i looked onto the table the only thing given was the Ka for C6H5COOH …
  8. Chemistry

    Suppose that 30.0 mL of 0.20 M C6H5COOH(aq) is titrated with 0.30 M KOH(aq) (a) What is the initial pH of the 0.20 M C6H5COOH(aq)?
  9. Chemistry

    Calculate the pH for each of the following points in the titration of 50.0 mL of a 2.7 M H3PO3(aq) with 2.7 M KOH(aq). pKa1 = 1.3 and pKa2 = 6.7 a) before addition of any KOH b) after addition of 25.0 mL of KOH c) after addition of …
  10. Chemistry

    Calculate the pH for each of the following points in the titration of 50.0 mL of a 2.7 M H3PO3(aq) with 2.7 M KOH(aq). (a) before addition of any KOH (b) after addition of 25.0 mL of KOH (c) after addition of 50.0 mL of KOH (d) after …

More Similar Questions