Question

A pH titration of a 300.0 mg sample of a weak acid HA in 100 mL of water is done by titrating with a 0.095M solution of NaOH. The resulting titration curve is shown below. Complete the following: (10 pts) a. The volume of NaOH used to titrate the solution to the endpoint is 15.5 ml, determine the number of moles of acid b. Determine the molar mass of the acid.

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Answer 1

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Tas 917 No.of moles of NaoH = Molarity X Volumele _ = p. 695 x 155 Xo? = 1047 x 103 mol Since, Valence factor of both NaOH & HA are 1. | No. of moles of Acid HA = Norot moles of 1. Naoh No.of moles of HA = 11:47x108mon b) No. of moles = of HA Los molar mass = mass in goom Molar mass 300 x 103 1.67 x 103 Molar mas = of Acid. 204.1 almell 5 92} Assign oxidation Number CIOs + 3H2O + 3 As 0 => 3 Asa + C1 + 6HF. 1+5 +3 +5 -1 les Oxidation No. of Aš increases from +3. ho + 5 w bile Oxidahon no. of cí I decreases from +5 to -1 @_Element oxidised → Oxidising Agent → As. 210g 1 bl Element Reduced a ci? fosscanne Reducing Agent Asos nei ALSACO 0

1103) Comparing experiment 1 4 2 data, [A] = constant. But cone of B' increases 4 time and Role increases by focher det, Rale = K[Aj[814 . of [4B789 = 2x Rete. aj.99 = '2 n 44 = 442 Similarly, from experinsest data 143 Be constant 2 A7? S.4x Rate loi 22 = 0 22=22 || a ro Order = x + y = 2+ = = b) Rate & KEAJ² [B]/ KE 0:035 Co2 1772 = from expen: I 1 data ] TK 3.5 volue of Rate constant e) Rote law is : Rare = k.CA]²[B]k] ed with Car Searter