Question

Chemistry 3A Worksheet #10.3 Group Name 1) Answer the following questions based on this balanced equation: H2SO4 + 2 NaOH => Na,SO4 + 2 H2O A) What is the concentration of a 10.00 mL sample of H2SO, if it takes 23.5 mL of 0.100 M NaOH to neutralize it? B) What volume of 0.125 M NaOH should be necessary to titrate a 5.00 mL sample of 1.0M H2SO4? 2) What mass of AgCl precipitates when 15.0 mL of 1.5 M AgNO, is mixed with excess Naci? AgNO3(aq) + NaCl(aq) => AgCl(s) + NaNO3(aq)

Answer 1

7 ₂ Sou + 2NaOH → Na₂SO4 + 2H₂O ay. Volume of H₂SO4 = 10.0mL = 0.01L Volume of NaOH = 23.5 mL = 0.0235 L lonen of NaOH = 0.1m = o.l mal, no. of male of NaOH = 0.0235 L x 0.1 mal, = 2.35 xlo mol Henie no. of mole of hou = fx no. of male of Nally = 1/2 x 2.35x 103 mod concentation of hoy = mole of H₂SO4 + x 2.358 10 mal Volume of H₂SO4 0.01L conen of H₂SO4 = 0.1175 m by Volume of NaOH = ? lonen of NaOH = 0.125 m - 0.125 mol/ Volume of H₂SO4 = 5.0ml = 0.005 L lonen of 4₂04 = 1.om = 1.0 mol/L no. of mole of hy soy = 0.00SLX 1.0 mol = 0.nos mol Hence no. of mole of NaOH = 2x no. of male of H₂SO4 = 2x 0.005 = 0.01 mol Volume of NaOH = no. of mole 0.01 mol con of NaOH 0.125 mol/L = 0.08L 80mL.

27 Volume of AgNO₃ = 15.0mL = 0.01SL Come of AgNO, & 1.5m = 1.5 mal/ no. of mole of AgNO₃ = 0.0ISLX 1.5 mol -0.0225 mol I mole A No, produce Imole of Age, then 0.0225 mol AgNO₃ produse 0.0225 mol Agle. no. of mole of Agul ppt = 0.0225 mol. mars of Aglul ppt = 0.0225 mol x 143.39 mod = 3.224 g. Неусе, 3.2244 of 40 Рpt .