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Posted by on Monday, May 2, 2011 at 3:20pm.

A solution is made by mixing exactly 500 mL of 0.178 M NaOH with exactly 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentrations of H+, CH3COOH, CH3COO-, OH-, and Na+.

  • Chemistry - , Monday, May 2, 2011 at 3:38pm

    The secret to solving these problems is to recognize what you have in solution at the point which the question addresses. Then write an equation and prepare an ICE table. Like the following which I will start but you must finish.
    500 mL x 0.178 = 89.0 mmoles NaOH
    500 mL x 0.100 = 50.0 mmoles CH3COOH
    .....NaOH + CH3COOH ==> CH3COONa + H2O
    initial 89.0..50.0........0.........0
    change..-50.0.-50.0......+50.0..+50.0
    equil...39.0...0..........50.0..50.0

    So you have 39.0 mmoles NaOH after the reaction; therefore, the pH will be determined by the OH^- from NaOH.
    (OH^-) = mmoles/mL
    mmoles = 39.0 and mL = 1000 mL
    (OH^-) = 0.039mols/L = 0.0390 M

    The others are done similarly.

  • Chemistry - , Monday, May 2, 2011 at 4:51pm

    I calculated all of them and this is what I got:

    1) [H+] = 2.56x10^-13 M

    2) [OH-] = .039 M

    3) [Na+] .089 M

    4) [CH3COO-] = .05

    5) [CH3COOH] = 0

    All of them were right except number 5 and I don't know what the answer to number 5 is suppose to be.

  • Chemistry - , Monday, May 2, 2011 at 5:32pm

    I would have agreed with all of your answers, including number 5, because whatever it is it must be so small as to be negligible. YOu might try this.
    CH3COOH ==> H^+ + CH3COO^-
    Ka = (H^+)(CH3COO^-)/(CH3COOH)
    If I substitute Ka = 1.8 x 10^-5 (your text may have a different number) and (H^+) from the problem as 2.56E-13 along with (CH3COO^-) = 0.05 from the problem and solve for (CH3COOH), I get something like 7.11E-10 which is, indeed, quite small. Or we could look at it another way, CH3COO^- is a base and hydrolyzes to produce
    CH3COO^- + HOH ==> CH3COOH + OH^-

    Kb = (Kw/Ka) = (1E-14/1.8E-5) = 5.55E-10 = (CH3COOH)(OH^-)/(CH3COO^-).
    If I substitute 0.039 for OH and 0.05 for CH3COO^- and solve for CH3COOH, I get 7.11E-10. Same value. That's a reasonable answer I think and it IS small. Let me know if this doesn't do it.

  • Chemistry - , Monday, May 2, 2011 at 5:39pm

    yes 7.11E-10 was the right answer thank you very much for the help :)

  • Chemistry - , Friday, March 20, 2015 at 6:27pm

    Wait, then why isn't the concentration of [H+] also considered to be zero since the value is so small?

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