Problems
1) Balance the following oxidation-reduction reaction:
Cu2+(aq) + Al(s) -> Al3+(aq) + Cu(s)
2) What volume of 0.185M NaOH solution gives 0.150 mol of NaOH?
3) Water is added to 30 mL of 0.875M KNO3 until its final volume reaches 500 mL. Calculate the final concentration.
4) A 0.800 L sample of industrial waste water is thought to be contaminated with Cu2+. An excess of Na2S is added to the drinking water and 0.0196 g of CuS is formed. What is the concentration of Cu2+ in our sample in grams/liter?
5) How many milliliters of 0.626M NaOH are needed to neutralize 30 mL of 0.250M H2SO4 solution? (HINT: write the balanced chemical equation for addition of NaOH and H2SO4 first)
Answers
1) 3Cu2+(aq) + 2Al(s) -> 2Al3+(aq) + 3Cu(s)
2) (1L/0.185 mol NaOH)*0.150 mol = 0.811 L or 811 mL
3) Mf = MiVi/Vf = (0.857M KNO3)(30 mL)/(500 mL) = 0.0525M KNO3
4) %Cu in CuS = (64g/mol/96g/mol)*100 = 66.7%
0.667*0.0196g = 0.0131 g
0.0131g/0.800L = 0.0164 g/L of Cu+
5) H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l)
0.030L*(0.250 mol H2SO4/L) = 0.00750 mol H2SO4
0.00750 mol H2SO4*(2 mol NaOH/1mol H2SO4) = 0.0150 mol NaOH
0.0150 mol NaOH*(1L/0.626 mol NaOH) = 0.0240 L = 24.0 mL