Chem 1 f92 assignment 5

1. Running head: CHEM 1F92 ASSIGNMENT 1
CHEM 1F92 Assignment 5. Reactions in Aqueous Solution; Solution Stoichiometry and
Analysis
1. (a) Write balanced molecular equations for the following potential precipitation reactions.
Indicate the states of reactants and products [(aq) or (s)]. (b) In those cases where a
precipitate forms, write the net ionic equation. If there is no reaction, state "No reaction."
(i) CoCl2 (aq) + K2S (aq) → COS (s) + 2 KCl(aq)
net ionic : Co2+ (aq) + S2- (aq)....> CoS(s)
(ii) 2 NaCl (aq) + (NH4)2S (aq) → No reaction
(iii) (NH4)2SO4 (aq) + Ba(NO3)2 (aq) → BaSO4 (s) + 2 NH4NO3 (aq)
net ionic: SO42- (aq) + Ba2+ (aq) ......> BaSO4(s)
(iv) CuCl2 (aq) + K2S (aq) → CuS(s) + 2 KCl (aq)
net ionic equation : Cu2+ (aq) + S2- (aq)....> CuS(s)
2. Each of the following salts can be prepared from an acid and a base. Write the balanced
molecular equation and the net ionic equation for the preparation of each. Indicate states of
the reactants and products [(aq), (s), (l)]. Review solubility rules if necessary to determine
the solubility of the reactants.
(a)
HClO4(aq) + KOH(aq) ....> KClO4 (aq) + H2O (l)
net ionic equation is H + (aq) + OH- (aq) ....>H2O (l)
(b)
H3PO4 (aq) + 2 CsOH (aq) ....> Cs2HPO4 (aq) 2 H2O(l)
net ionic equation is H + (aq) + OH- (aq) ....>H2O (l)
(c)
Ca(OH)2(s) + 2 HNO3 (aq).....> Ca(NO3)2 (aq) + 2 H2O(l)
net ionic equation is H + (aq) + OH- (aq) ....>H2O (l)
3. 3.3872 g of Na4SiO4 = 3.3872 / 184.042 = 0.018 mole.
(a) molarity of Na4SiO4 = 0.018 * 1000 /100 = 0.18 M
(b) molarity of the sodium cation = 4 *0.18 = 7.2 M
(c) the molarity of the silicate anion = 0.18 M
4. You have a solution of 0.1127 M calcium chloride.
(a) How many moles of solute are contained in 27.85 mL of solution = 0.02785 * 0.1127 =
0.00314 mole
(b) grams of solute are contained in 27.85 mL of solution = 0.00314 *110.98 = 0.348 gm
(c) volume (in millilitres) of solution is needed to obtain 0.00650 moles of solute:
27.85*0.00650 / 0.00314 = 57.65 ml
5.Suppose you need 2.50 L of 1.60 M sulfuric acid, but only the concentrated acid (17.8 M)
is available. What volume (in L) of the concentrated acid must you dilute to 2.50 L to obtain
1.60 M sulfuric acid?
volume (in L) of the concentrated acid = 2.50 * 1.60 / 17.8 = 0.2247 L = 224.7 ml.