Question

Now to the ocean! I was scuba diving and gazed around at the underwater world, remembering that the ocean was just a huge solution, filled with solutes. 13) Give an example of a solute present in ocean water. Explain what a “solute” is in your answer and how your example fulfills this requirement. 14) Welding is be done in the ocean water! Many times, as welding occurs, a cloudy substance would form around the area. Welding uses a compound that has lead(II) nitrate in it. I knew that ocean water has sodium chloride in it. a) Type the balanced molecular equation for the formation of the cloudy material, including all phase labels. b) Type the complete and net ionic equation for the formation of this cloudy material. 15) I also remembered from general chemistry that there was gold in the ocean. If a very small amount was floating around in the form of Gold(I) sulfate, maybe I could use the principle of oxidation-reduction to recover it? a) Choose a metal from the activity series, that will replace gold from gold(I) sulfate to elemental gold and write a balanced reaction to illustrate this transformation, including all phase labels b) Determine the oxidation number of each element and write it below each-reactant and product in your molecular equation above. c) Identify which reactant has been oxidized and which reactant has been reduced. Explain how could you tell? 16) In chemistry class we also studied concentration. We took a 100-mL sample of ocean water and evaporated it to dryness. The result was 3.00 grams of NaCl and 2.00 grams of MgCl2. a) Give the individual balanced equation for the dissolving of MgCl2 in water to form ions, including phase labels. b) What is the concentration, in molarity, of only the magnesium ion (Mg+2), in the 100-ml sample,? 17) I also remembered using a “titration” to determine concentrations. I knew about acid-base titrations, but what about ions? I knew there was also chloride in the ocean. Could I recover it with titration…Consider the following: AgNO3( )+ MgCl2( ) + NaCl( )→AgCl( )+Mg(NO3)2( ) + NaNO3( ) a) Fill in the phase labels for each reactant and product above and balance the equation. Put your phase labels in bold and your coefficients in bold. b) I took another 100.0 ml sample of seawater, which contained the salts listed in the reactants and titrated them with silver nitrate. I isolated 13.63 grams of silver chloride. What must have been the molarity of the chloride ions in my seawater sample.

Answer 1

13. A solute is a substance which dissolves in solvent to make a solution. In ocean water sodium chloride is present which dissolves in ocean water to make a solution. So, sodium chloride is solute present 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 and appears as a cloudy substance.

b) Compelet ionic equation :

Pb⁺² + 2 NO₃^- + 2 Na⁺ + 2 Cl^- = PbCl₂ + 2 Na⁺   + 2 NO₃^-

Net ionic equation:

Pb⁺² + 2 Cl^- = PbCl₂

15 . A redox reaction can be used to recover gold grom gold sulphate. If you use a metal which in more reactive than gold in reaction with gold sulphate then gold (metal) will come out from the 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 mole of 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) = 0.21 M