It takes15.8mL of 0.10M NaOH standard solution to reach an equivalence point for

Question

It takes15.8mL of 0.10M NaOH standard solution to reach an equivalence point for a titration of 20.00 mL of a weak monoprotic acid,

(1) What is molarity of this acid?



(2) The initial pH of the acid before adding any base was 3.48. What is Ka value of this acid?



(3) What is % Inozation of the acid at the beginning?




(4) What is pKa value of this acid?



(5) If the true value of the pKa is 6.07. What is the absolute error?



(6) What is the relative % error in measured pKa value?

(7) pH value at the mid-point (half-way point) to the equivalence point found to be 5.72. What is pKa according to this mid point pH value?



(8) What is the absolute error in pKa according to pKa from the mid-point pH value? The true value is given in # (5)



(9) What is the relative % error in pKa according to pKa from the mid-point pH value?

Explanation / Answer

(a)first find the # of moles of NaOH.

since M = moles/Litres, what you do here is divide 15.8 ml by 1000 to get your volume units in Litres.

once you have the volume in litres, you multiply it by 0.10 M ( moles/litres ) thus cancelling the litres and giving you # moles of NaOH.

once that is done you can do a 1;1 stoichiometric ratio for your moles of OH- ( NaOH ) and H+ ( HCl ), giving you the moles of H+

and finally you divide the volume of HCl by 1000 to give you the volume units in litres and put moles of litres to find the molarity of Acid



15.8/1000 = 0.0158 L x 0.10 moles/litres = 0.00158 moles

0.00158 moles of OH- x 1 mole of H+/ 1 mole of OH- = 0.00158 moles of H+

20ml/1000 = 0.020 litres

0.00158 moles/ 0.020 litres = 0.079 M of acid

(b) [H+] = 10^-ph = 3.31 * 10^-4

Ka = 3.31*10^-4 * .10/.079 = 4.19 * 10^-4

(d) pKa = -log[ka] = 3.37

(e) error = 6.07-3.37 = 2.69

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