Question

4. Acid-base reaction is a neutralization reaction. If 12.50 mL of 1.20 M NaOH solution is used to neutralize 45.25 mL of HNO3 solution, calculate the concentration of the original HNO3 solution.

   (A). 0.231 M

   (B). 0.995 M

   (C). 0.887 M

   (D). 0.331 M

   (E). 9.90 M

5. Acid-base reaction is a neutralization reaction. If 12.50 mL of 1.20 M NaOH solution is used to neutralize 45.25 mL of H2SO4 solution, calculate the concentration of the original H2SO4 solution. 2NaOH + H2SO4  Na2SO4 + 2H2O (balanced?)

   (A). 0.231 M

   (B). 0.995 M

   (C). 0.887 M

   (D). 0.166 M

   (E). 0.331 M

10. Milk of magnesia contains magnesium hydroxide. What mass of Mg(OH)2 is needed to prepare 500.00 mL of 0.0500 M solution of magnesium hydroxide?

   (A). 0.987 g of Mg(OH)2

   (B). 0.654 g of Mg(OH)2

   (C). 1.46 g of Mg(OH)2

   (D). 10.7 g of Mg(OH)2

   (E). No correct answer is given

13. Write the complete molecular equation for the reaction between Na2SO4 and BaCl2 solutions.

14. Write the complete ionic equation for the reaction between Na2SO4 and BaCl2 solutions.

15. Write the net ionic equation for the reaction between Na2SO4 and BaCl2 solutions.

16. Identify the driving force(s) in the reaction between Na2SO4 and BaCl2 solutions.

Answer 1

4. We need to remember that at equilibrium, the moles of acid = moles of bases added.

We start with 12.50 mL (= 0.0125 L ) of 1.20 M NaOH solution, so that moles of base added = (0.0125 L)(1.20 mol/L) = 0.015 mol NaOH. At equilibrium, this is equal to the number of moles of HNO₃ neutralized. The volume of HNO₃ added is 45.25 mL (=0.04525 L).

Hence, molarity of HNO₃ = 0.015 mol/0.04525 L = 0.331 M

Ans: (D) 0.331 M

5. In this case also, we employ the same idea as above. As before moles of NaOH added = 0.015. Now, as can be seen from the reaction, the molar ratio of NaOH:H₂SO₄ is 2:1, i.e, 2 moles of NaOH neutralize 1 mole of H₂SO₄.

Therefore, 0.015 mol NaOH will neutralize (0.015/2) = 0.0075 mol H₂SO₄.

The volume of H₂SO₄ added is 45.25 mL (=0.04525 L).

Hence, molarity of H₂SO₄ is (0.0075 mol/0.04525 L) = 0.166 M

Ans: (D) 0.166 M

10. We first need to find out the gm-molar mass of Mg(OH)₂ from a knowledge of the atomic masses of the elements. The gm-molar mass of Mg(OH)₂ is {24.3 + 2.(15.99 + 1.008)} = 58.296 gm/mol

Now, amount of Mg(OH)₂ present in 500 mL (=0.5 L) of 0.0500 M milk of magnesia is (0.5 L)(0.0500 mol/L) = 0.025 mol

Hence, amount of Mg(OH)₂ added to obtain the milk of magnesia solution is (58.296 gm/mol)(0.025 mol) = 1.4574 gm.

Option (C) is the closest option, hence the answer is 1.46 gm.

Ans: (C) 1.46 gm of Mg(OH)₂

13. Na₂SO₄ + BaCl₂ ----------> BaSO₄ ↓ + 2NaCl; BaSO₄ appears as a white precipitate.

14. Both Na₂SO₄ and BaCl₂ are ionic compounds and exist as ions in solution. Hence, we can write down the complete ionic equation as

2Na⁺SO₄^2- + Ba²⁺2Cl^- -------------> BaSO₄ ↓ + 2 Na⁺Cl^-

15. The net ionic equation is the formation of BaSO₄ precipitate; the reaction is

Ba²⁺ + SO₄^2- --------> BaSO₄ ↓

16. The driving force for the above reaction is the formation of BaSO₄ . BaSO₄ is insoluble in aqueous medium (the reaction takes place in aqueous or slightly acidic medium as all both the starting materials are soluble), hence it separates out from the medium as a white solid. If the reaction is conducted under equilibrium conditions, then. according to Le Chatelier principle, removal of one of the products will favour the forward reaction. Hence, more reaction will occur giving more of the solid precipitate.