Now to the ocean! I was scuba
Now to the ocean! I was scuba diving and gazed around at theunderwater world, remembering that the ocean was just a hugesolution, filled with solutes. 13) Give an example of a solutepresent in ocean water. Explain what a “solute” is in your answerand how your example fulfills this requirement. 14) Welding is bedone in the ocean water! Many times, as welding occurs, a cloudysubstance would form around the area. Welding uses a compound thathas lead(II) nitrate in it. I knew that ocean water has sodiumchloride in it. a) Type the balanced molecular equation for theformation of the cloudy material, including all phase labels. b)Type the complete and net ionic equation for the formation of thiscloudy material. 15) I also remembered from general chemistry thatthere was gold in the ocean. If a very small amount was floatingaround in the form of Gold(I) sulfate, maybe I could use theprinciple of oxidation-reduction to recover it? a) Choose a metalfrom the activity series, that will replace gold from gold(I)sulfate to elemental gold and write a balanced reaction toillustrate this transformation, including all phase labels b)Determine the oxidation number of each element and write it beloweach-reactant and product in your molecular equation above. c)Identify which reactant has been oxidized and which reactant hasbeen reduced. Explain how could you tell? 16) In chemistry class wealso studied concentration. We took a 100-mL sample of ocean waterand evaporated it to dryness. The result was 3.00 grams of NaCl and2.00 grams of MgCl2. a) Give the individual balanced equation forthe dissolving of MgCl2 in water to form ions, including phaselabels. b) What is the concentration, in molarity, of only themagnesium ion (Mg+2), in the 100-ml sample,? 17) I also rememberedusing a “titration” to determine concentrations. I knew aboutacid-base titrations, but what about ions? I knew there was alsochloride in the ocean. Could I recover it with titration…Considerthe following: AgNO3( )+ MgCl2( ) + NaCl( )→AgCl( )+Mg(NO3)2( ) +NaNO3( ) a) Fill in the phase labels for each reactant and productabove and balance the equation. Put your phase labels in bold andyour coefficients in bold. b) I took another 100.0 ml sample ofseawater, which contained the salts listed in the reactants andtitrated them with silver nitrate. I isolated 13.63 grams of silverchloride. What must have been the molarity of the chloride ions inmy seawater sample.
Answer:
13. A solute is a substance which dissolves in solvent to make asolution. In ocean water sodium chloride is present which dissolvesin ocean water to make a solution. So, sodium chloride is solutepresent in ocean wtaer.
14 a) Reaction between sodium chloride and lead nitrate:
Pb(NO3 )2 (aq) + 2 NaCl (aq)= PbCl2 (s) + 2 NaNO3 (aq)
PbCl2 (lead chloride) is insoluble in water andappears as a cloudy substance.
b) Compelet ionic equation :
Pb+2 + 2 NO3– + 2Na+ + 2 Cl– = PbCl2 + 2Na+ + 2 NO3–
Net ionic equation:
Pb+2 + 2 Cl– = PbCl2
15 . A redox reaction can be used to recover gold grom goldsulphate. If you use a metal which in more reactive than gold inreaction with gold sulphate then gold (metal) will come out fromthe reaction.
a) One such metal is Magnesium (Mg)
Reaction: Mg (s) + Au2SO4 (aq) = 2 Au (s)+ MgSO4 (aq)
b) Mg0 (s) ++1Au2SO4 (aq) = 2 Au0(s) + Mg+2SO4 (aq)
c) Mg (metal) loses electrons to form Mg+2. So, Mg is oxidised.And Au+1 accepts electrons to form Au(0). Au+ is reduced.
16. a) MgCl2 + H2O = Mg+2(H2O) + 2 Cl- (H2O)
b) there is 2.00 grams og MgCl2 in 100 ml. ocean water
2.00 grams of MgCl2 = (2.00/ 95.211) moles of MgCl2 = 0.021 moleof MgCl2 (molar mass of MgCl2 is 95.211 g/ mole)
Now, 0.021 mole MgCl2 contains 0.021 mole Mg+2
So, 100 ml. of water contains 0.021 mole Mg+2
Concentration of Mg+2 in water = (0.021 * 1000)/ 100 (M) = 0.21M