Question

1. A certain experiment requires 624.0 mL of 0.00445 M NaOH.

i). Calculate the number of grams of NaOH required to make this solution.

ii). Calculate the mass percentage of a 0.00445 M NaOH solution (the density of solution is 1.52 g/mL).

iii). Calculate the pOH and pH of this 0.00445 M NaOH solution.

iv). What volume of 0.175 M NaOH solution is required to make 624.0 mL of 0.00445 M NaOH solution?

v). Calculate the pOH and pH of the following mixture: 43.7 mL of a 0.18 N H2SO4 solution after it is mixed with 57.3 mL of a 0.35 N NaOH solution.

Answer 1

Volume of NaOH (V.) = 624.0 ml Molarity of NaOH (M) = 0.00445 M (i) moles of NaOH required (n) = Mix V, Lin litres). = 0.00445 x 624.0 1000 = 2.78 x 10-3 mol. mass of NaOH required (mi) = moles x Molar mars = 2.781 100 lux 40 glmol = 10112 x10 g cil density of solution (d) = 1.52 g/m2 mass of Solution (m) - dxv, . = 1.52 X 624.0 c.948.489 mass percentage = mass of Naon x 100 mass of solution 1.112 x 1024 x 100 948.48 = 0.0117 % (ii) Naotlaq - Nat(aq) +ou-leq). [04-] = 0.00445 M pOH = -log [OH-] = 2.35 pH = pkw-poh = 14-2.35 = 11.65

0. (iv) Let volume of Naot required = V me. - New molarity (M') = M xv - 624 ~ 0.0045 = 0.175XV 624 V16.0ML Ivo 2 Naon + H2SO4 - Na, 504 + 2H20. milliequivalents of NaOH = 0.35 N *57.3 mL = 20.055 meq. milliequivalents of H2SO4 = 0.18 Nx 43.7 ml = 7.866 meq. 6 meq..... NaOH is in excess. thenie milli Hence, nequivalents of - 20.055 - @ 7.866 Naon left - 12.189 meq. Naon - Nat cap) + OH (as) [OH-] = moles of Noon rofume of solution litres) 12.189 x10-3 (57.3+43.77X10-3 poha -log(OH] = -log(0.121) - 0.92 pH = pkw-poh - 14-09.92 = 13.08 = 0.121 M