chemistry

If 25.00 mL of 0.20 mol/L HCO2H(aq) is titrated with 0.20 mol/L NaOH(aq) (the titrant), determine the pH
a) befor titration began

b)after 10.00 mL of NaOH has been added

c) at the equivalence point

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asked by Bradely
  1. a) before titration begins. That is a 0.2 M solution of HCOOH.
    HCOOH ==> H^+ + HCOO^-

    (HCOOH) = 0.2-x
    (H^+) = x
    (HCOO^-) = x
    Ka = (H^+)(HCOO^-)/(HCOOH)
    Plug the above variables into Ka and solve for (H^+), then convert to pH.

    b) after 10.00 mL NaOH has been added.
    HCOOH + NaOH ==> HCOONa + H2O

    mols HCOOH initially = M x L = ??
    mols NaOH added = M x L = ??
    mols HCOONa formed = ??

    pH = pKa + log[(base)/(acid)]

    c)at the equivalence point. We have the salt HCOONa + H2O at the equivalence point.
    HCOO^- + HOH ==> HCOOH + OH^-
    Kb = Kw/Ka = (HCOOH)(OH^-)/(HCOO^-)
    (HCOOH) = x
    (OH^-) = x
    (HCOO^-) = 0.2 - x
    Plug into Kb and solve for x = (OH^-), then convert to pH.

    Post your work if you get stuck.
    Check my thinking. Check my work. Check my arithmetic.

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    posted by DrBob222
  2. a) Ka = (H^+)(HCOO^-)/(HCOOH)
    = (x)(x)/(0.2-x)
    = (x^2) / 0.2 - x

    Next step please...I Hate these problems :P:)

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    posted by Sarah

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