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    1.00g of potassium ethanoate is dissolved in 50 cm^3 of 0.200 mol dm^-3 ethanoic acid (Ka = 1.74 * 10^-5 mol dm^-3)

    Find the pH.

    When you work it out could you please show me step by step how you did it, I know how to use Ka calculations but this one gave me an answer of -1.232 for the pH. ????????????????
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    1) Find the number of mol or potassium ethanoate
    2) use the equation, pH = pKa + lg (concentration of salt/ concentration of acid)
    - in this case, concentration of salt is the no. Of mol per dm3 of potassium ethanoate
    - acid ... just find the no. Of mol (mol = concentration x volume)
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    Correct me if I am wrong! Thanks
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