Question

# when 75.0 ml of .150 m NaCl solution and 75.0mL of a 0.250 M Pub(NO3)2 are...

when 75.0 ml of .150 m NaCl solution and 75.0mL of a 0.250 M Pub(NO3)2 are mixed, a white precipitate forms.

a. Identify the precipitate in the reaction.

b. write out the balanced molecular equation and net ionic equation for the reaction.

c. Calculate the mass (in g) of precipitate formed.

d. calculate the concentration s of the remaining ions in solution. (what is the total volume after the solutions are mixed? How many moles of each ion remain in solution?)

a)

The reaction:

according to solubility rules, PbCl(s) is not soluble, will form precipitate

b)

Pb(NO3)2(aq) + NaCl(aq) = PbCl(s) + NaNO3(aq)

balance molecular

Pb(NO3)2(aq) + 2NaCl(aq) = PbCl(s) + 2NaNO3(aq)

the complete ionic:

Pb+2(aq)+ 2NO3-(aq) + 2Na+(aq)+ 2Cl-(aq) = PbCl(s) + 2Na+(aq)+ 2NO3-(aq)

net ionic (get rid of spectator ions, NO3- and Na+)

Pb+2(aq)+ 2Cl-(aq) = PbCl(s)

c)

mass of preicpitate formed

mmol of Cl- = MV = 0.15*75 = 11.25 mmol of Cl-

mmol of Pb+2 = MV = 0.25*75 = 18.75 mmol of Pb+2

raito is 1:2 so

11.25 mmol of Cl- --> 1/2*11.25 = 5.625 mmol of Pb+2 required

mmol of PBCl2 formed = 5.625 mmol

mass = mmol*MW = 5.625*278.1 =1564.3125 mg = 1.564 g of PbCl2(s) forms

d)

V total final = 75+75 = 150 mL

Get

[Cl-] = 0

[Pb+2] = 18.75 -5.625 / V = 13.125 / 150 = 0.0875 M

[Na+] = mmol/V = 11.25 /150 = 0.075 M

[NO3-] = 18.75 *2 /150 = 0.25 M

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