A titration involves adding a reactant of known quantity to a solution of an ano

Question

A titration involves adding a reactant of known quantity to a solution of an another reactant while monitoring the equilibrium concentrations. This allows one to determine the concentration of the second reactant. The equation for the reaction of a generic weak acid HA with a strong base is HA(aq)+OH(aq)A(aq)+H2O(l) A certain weak acid, HA, with a Ka value of 5.61×106, is titrated with NaOH. Part A A solution is made by titrating 9.00 mmol (millimoles) of HA and 3.00 mmol of the strong base. What is the resulting pH? Express the pH numerically to two decimal places. pH = SubmitHintsMy AnswersGive UpReview Part Part B More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 57.0 mL ?

Explanation / Answer

Part A. HA(aq)+   OH(aq)A(aq) + H2O(l)

        I   9 mmol    3mmol       0

       C -3 mmol   -3mmol +3mmol

       E 6 mmol    0              3 mmol

Since, strong base is assumed to be reacting all of it, so after addition, according to Henderson Hasselbach equation for buffer made of weak acid HA and its salt/conjugated base A-,

pH= pKa + log [A-]/[HA] i.e. pH = -log(Ka) + log (3mmol/V)/(6 mmol/V)

i.e. pH = -log(5.61×106) + log (3mmol/V)/(6 mmol/V) i.e. pH = 4.95

Part B. At equivalence point, [A-] = [HA]

So, pH= pKa + log [A-]/[HA] i.e. pH = pKa = -log(Ka) = -log(5.61×106) = 5.25

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