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    • Thread Starter

    1. 20 cm3 of methanoic acid (Ka = 1.8 x 10-4 moldm-3) of concentration 0.10 moldm-3 is titrated against sodium hydroxide of concentration 0.05 moldm-3.
    Calculate the pH of the solution after 10 cm3 of the alkali has been added

    I know how to get the answer, I'm just confused in understanding the context.

    HCOOH = H+ + HCOO-
    Initially, we have 0.002 mols of HCOOH and 0 mols of the other two
    When we add NaOH of 0.0005 mols, we have 0.0015 mols of HCOOH and 0.0005 mols of the other two
    Now, instead of using Ka to work out concentration of H+, why can we not just 0.0005 mols and work out its concentration?
    Or why can we not use [H]^2 when working out Ka?

    Two questions

    Hi! I hope that you know you are actually calculating the pH of a buffer solution instead of just a pure solution of acid or base alone. You have to take into account that the strong conjugate base will also hydrolyse to produce hydroxide ions in the process, which then affect the pH of the solution as well.
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