11. Consider the titration of equal volumes of ic acid) with 0.1 M NaOH. W (A) (
ID: 557842 • Letter: 1
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11. Consider the titration of equal volumes of ic acid) with 0.1 M NaOH. W (A) (B) (C) hich of the following would be the same for both titrations? the initial pH the pH at the half-equivalence point the pH at the equivalence point (E) all of the above 12. Given the following data collected from the titration of a 10.00 mL sample of an unknown weak monoprotic acid solution with 0.1500 M KOH, what is the Ka of the weak acid? volume of base added (mL) 0.00 2.50 5.00 7.50 10.00* 15.00 solution pH 2.97 4.64 5.12 5.60 9.00 12.48 The endpoint of the titration, as determined by phenolphthalein. (A) 7.6 × 10-6 (B) 1.3 x 10-9 (C) 1.0 x 10-9 (D) 2.5 × 10-6 (E) 2.3 × 10-5 13. The expression for the solubility product of Ag,PO, is: (C) [3 × Ag+]"po,31 (E) [Ag+] 3 [PO43-] +13 PO4 (A) [Ag+]3 [P043-] (B) [Ag+]3 [Po /A&P;Explanation / Answer
OPTION A. From the concentration of KOH and the volumes of the base and acid taken, we can calculate the concentration of the weak acid present as per law of equivalence to give Ma = (VbMb)/Va => Ma = (10x0.15)/10 = 0.15M.
The expression for Ka is given as [H+][A-]/[HA] for a weak acid HA with conjugate base A-. We know that the concentration of acid is 0.15M. At half the equivalence point, the pH is considered equal to the pKa of the acid as per Henderson-Hasselbach equation as that is when the concentration of protons is equal to the concentration of salt KA (for acid HA). Thus, using the pH at half-equivalence point of 5mL which is 5.12, we can put concentration of protons as [H+] = 10-pH = Ka which gives 10-5.12 = 7.6x10-6.
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