Science - chemistry

Calculate the pH of a solution prepared by adding 36 cm3 of 0.15 M methanoic acid to 27 cm3 of 0.20 M NaOH.�(Ka for methanoic acid = 1.77 x 10-4)

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asked by Teagan
  1. See your other post.
    mmols HCOOH = mL x M = estimated 5.4
    mmols NaOH = mL x M = estd 5.4

    ......HCOOH + NaOH ==> HCOONa + H2O
    I.......5.4.....0........0.......0
    add............5.4...............
    C.......-5.4..-5.4.....5.4........
    E.......0.......0.......5.4

    So this is not a buffer problem because you have no acid/base mixture. The base is there (5.4 mmols HCOONa) but the acid was neutralized; therefore, the pH is determined by the hydrolysis of the salt (HCOONa)
    (HCOONa) = 5.4 millimols/63 mL volume = 0.086M
    ........HCOO^- + HOH ==> HCOOH + OH^-
    I.......0.086..............0......0
    C.........-x...............x......x
    E......0.086-x.............x......x

    Kb for HCOONa = (Kw/Ka for HCOOH) = (x)(x)/(0.086-x)
    Solve for x = OH^- and convert to pH.

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    posted by DrBob222

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