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13 July, 11:55

Calculate the ph of a solution containing 0.0451 m potassium hydrogen tartrate and 0.028 m dipotassium tartrate. The ka values for tartaric acid are 9.20 * 10-4 (ka1) and 4.31 * 10-5 (ka2).

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  1. 13 July, 13:49
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    Given buffer:

    potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6)

    [KHC4H4O6] = 0.0451 M

    [K2C4H4O6] = 0.028 M

    Ka1 = 9.2 * 10^-4

    Ka2 = 4.31*10^-5

    Based on Henderson-Hasselbalch equation;

    pH = pKa + log [conjugate base]/[acid]

    where pka = - logKa

    In this case we will use the ka corresponding to the deprotonation of the second proton i. e. ka2

    pH = - log Ka2 + log [K2C4H4O6]/[KHC4H4O6]

    = - log (4.31*10^-5) + log [0.0451]/[0.028]

    pH = 4.15
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