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    ok how would i work out the Ka of ethanoic acid when given:

    1 M ethanoic acid which has a pH of 2.37


    also (i think im bein stupid here) but what would the ionic equation be when sulphuric acid reacts with potassium hydroxide
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    because ethanoic is a weak acid HA <---> H+ + A- we say that [H+] = [A-]

    so that for the equilibrium above Ka = [H+][A-]/[HA] = [H+]2/[HA]

    so knowing [HA] = 1 M and the pH = -log [H+] = 2.37 we can find [H+] and hence Ka

    H2SO4 + 2KOH ---> 2H2O + K2SO4

    which when removing the spectator ions SO42- and K+

    we get OH- + H+ ---> H2O
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    wow thanks for such a quick and informative answer, you've really helped me
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    (Original post by sharp357)
    wow thanks for such a quick and informative answer, you've really helped me
    welcome
 
 
 
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