Question

0.5 pts Question 1 A student titrates 0.139 g of an unknown monoprotic weak acid to the equivalence point with 44.6 mL of 0.100 M NaOH (aq). What is the molar mass of the weak acid? O 0.0615 g/mol 0.0312 g/mol 13.9 g/mol 31.2 g/mol O 61.5 g/mol

Answer 1

Molality of a solution is the ratio of the number of moles of solute to the Mass of solvent in kilogram. The unit of molality is mol/kg or m. Expression for molality: Molality Moles of solute Mass of solvent in Kg Given that Mass of acid (solute)=0.139 g Volume of NaOH = 44.6 mL Molarity of NaOH=0.100 M Consider the reaction, HA (aq) + NaOH(aq) NaA (aq) +H20(1) Calculate the number of moles of NaOH, Moles of NaOH = Volume of NaOH Xmolarity of NaOH = 44.6 prl x_11_xo 100 mol OPAL *1000 zł 0.100 ml = 0.00446 mol Calculate the number of moles of HA, Mass of HA Moles of HA=- Molar mass of HA _0.139g МА At equivalence point, the number of moles of HA will be equal to the number of moles of NaOH. Moles of HA = Moles of NaOH 0.139g = 0.00446 mol MHA 0.139g MHA 0.00446 mol = 31.2-8 mol Hence, the molar mass of monoprotic weak acid is 31.2

Given that, Concentration of pentanoic acid = 0.120 M K=1.45x10 Consider the reaction, CH, (CH), COOH(aq)+H,0(1) =CH (CH), COO (aq)+H30* (aq) Initial concentration (in M) 0.120 Change in concentration At equilibrium (0.120 - ) At equilibrium, the concentration of CH, (CH), COO and H,0* are same. Concentration of 1,0* = concentration of CH, (CH), COO = x The K, for the reaction is: KE CH, (CH), COOH,0*] (CH(CH), COOH] 1.45x10-5-_(x) (0.120 - x) x=1.31x10 M Hence, the concentration of H,0* is 1.31x10'M- Use the following relation to calculate the concentration of OH® ion, [H,0*][OH-]=10-44 [OH 7= 10-44 H,0*] 10-14 1.31x107 =7.63x10-12 M Hence, the concentration of OH ion is 7.63x10-12 M.