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    Question:
    A solution of phenol in water has a concentration of 4.7 g dm–3.
    (ii) Calculate the pH of this solution of phenol.

    Answer:
    Mr C6H5OH = 94 (1)
    [C6H5OH(aq)] 4.7/94 = 0.050 mol dm–3 (1)
    1.3 × 10–10 ≈ [H+(aq)]2 / 0.050 mol dm–3 (1) (‘=’ sign is acceptable) [H+] = √{(1.3 × 10–10) × (0.050) } = 2.55 × 10–6 mol dm–3 (1)
    pH = –log[H+] = –log 2.55 × 10–6 = 5.59 (1)
    3 marks: [H+]; pH expression ; calc of pH from [H+]


    where does the 1.3 *10^-10 come from

    i understand that this is [h+]^2 but there is not pKa or Ka value to calculate this from
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    (Original post by esmeralda123)
    Question:
    A solution of phenol in water has a concentration of 4.7 g dm–3.
    (ii) Calculate the pH of this solution of phenol.

    Answer:
    Mr C6H5OH = 94 (1)
    [C6H5OH(aq)] 4.7/94 = 0.050 mol dm–3 (1)
    1.3 × 10–10 ≈ [H+(aq)]2 / 0.050 mol dm–3 (1) (‘=’ sign is acceptable) [H+] = √{(1.3 × 10–10) × (0.050) } = 2.55 × 10–6 mol dm–3 (1)
    pH = –log[H+] = –log 2.55 × 10–6 = 5.59 (1)
    3 marks: [H+]; pH expression ; calc of pH from [H+]


    where does the 1.3 *10^-10 come from

    i understand that this is [h+]^2 but there is not pKa or Ka value to calculate this from
    It is the Ka of phenol
 
 
 
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