Question

How many grams of each of the following substances will dissolve in 100 mL of cold water? Consult a handbook or the Internet and cite your source. Ce(IO_3)_4 RaSO_4 Pb(NO_3)_2 (NH_4)_2SeO_4 Which of these substances is least soluble on a gram-per-100 mL basis? Suppose you have a solution that might contain any or all of the following cations. Cu^2+, Ag^+, Ba^2+, and Mn^2+. The addition of HBr causes a precipitate to form. After the precipitate is filtered off, H_2SO_4 is added to the supernate, and another precipitate forms. This precipitate is filtered off, and a solution of NaOH is added to the supernatant liquid until it is strongly alkaline. No precipitate is formed. Which ions are present in each of the precipitates? Which cations are not present in the original solution? Write balanced net ionic equations for the reactions, if any, that occur between Fe_2S_3(s) and HBr(aq) K_2CO_3(aq) and Cu(NO_3)_2(aq), Fe(NO_3)_2(aq) and HCl(aq), and Bi(OH)_3(s) and HNO_3(aq).

Answer 1

2a (1) Ce(IO4)4 :

Solubilty of cerium periodate is 2.72 - 2.88 mol/dm3 = 2.72 - 2.88 mol/litre = 2.72 - 2.88 mol/1000 mL

So for 1000 mL 2.72-2.88 moles can be dissolved

for 100 mL 0.272-0.288 moles i.e 0.272-0.288 moles/100 mL

So 0.288 x M wt /100 mL

0.288 x 839.72 g/100 mL

= 241.83 g/100 mL

so for 100 mL of cold water 241.83 grams can be dissolved.

2. RaSO4

The density of RaSO4 is 5.42 g/cm3.= 5.42 g/mL

So for 100 mL of cold water = 5.42 x 100 = 542 g/100 mL

3. Pb(NO3)2 density = 4.53 g/mL

So for 100 mL of cold water = 4.53 x 100 = 453 g/100 mL

4. (NH4)2SeO4

density = 1.769 g/cm³ (20 °C) = 1.769 g/mL

So for 100 mL of ice cold water = 176.9 g/100 mL can be dissolved.

2b (NH4)2SeO4 is least soluble on gram per 100 mL basis as it has least solubility/100 mL i.e 176.9 g/100 mL

3 Cu2+, Ag+, Ba2+, and Mn2+ & with addition of HBr only Ag+ and Ba+ will precipitate out as follows:

Ag+ + HBr ===> AgBr(s) + H+ On addition of HBr

Ba2+ + H2SO4 ===> BaSO4(s) + 2H+ On addition of H2SO4

Ag+ and Ba+ are present in each of the precipitates. Cu2+ and Mn2+ are not present in the original solution because if present, they would have formed Mn(OH)2(s) and/or Cu(OH)2(s).

4 a molecular: Fe2S3(s) + 6 HBr(aq) → 3 H2S(g) + 2 FeBr3(aq)

total ionic: Fe2S3(s) + 6 H(+) + 6 Br(-) → 3 H2S(g) + 2 Fe(3+) + 6 Br(-)

net ionic: Fe2S3(s) + 6 H(+) → 3 H2S(g) + 2 Fe(3+)

b CuCO3 is an insoluble salt so the ionic equation is

Molecular equation:  Cu(NO3)2 (aq) + K2CO3 (aq) --> CuCO3 (s) + 2KNO3 (aq)

Total ionic equation: Cu2+ (aq)+ 2 NO3- (aq)+ 2 K+ (aq)+ CO32-(aq) = CuCO3 (s)+ 2 K+ (aq) + 2 NO3- (aq)
  
Net ionic equation: Cu2+ (aq) + CO32- (aq)= CuCO3 (s)

c No reaction because there is a reaction only if a product is insoluble, a gas, or a nonelectrolyte but all the above substances are soluble electrolytes, so there is no reaction.

d Molecular equation: HNO3 (aq) + Bi(OH)3 (s) ----> Bi(NO3)3 (aq) + 3 H2O (l)

Net Ionic Equation : 3 [ H+ ] (aq) + Bi(OH)3 (s) ----> [ Bi3+ ] (aq) + 3 H2O (l)