Chemistry Concepts Review

Chemical Reactions and Precipitate Formation

Unfinished Reaction Analysis

The given unfinished reaction is:
Pb(NO3)2(aq) + Na2SO4(aq)
ightarrow (?)(s) + NaNO_3(aq)
From this equation, we need to deduce:

1. Identifying the Precipitate:
   • The options for the precipitate formed from this double displacement reaction include:
     • a. $Pb2SO4$
     • b. $NaNO_3$
     • c. $PbSO_4$
     • d. $Pb(SO4)2$
   • The correct answer is c. $PbSO_4$, which is a known insoluble compound, indicating that it will precipitate out of the solution.

Balancing the Reaction

2. Coefficients in the Complete Balanced Equation:
   • The balancing of the equation is necessary to reflect the law of conservation of mass. The possible options for coefficients are:
     • a. 1,1,1,1
     • b. 1,1,1,2
     • c. 1,2,1,1
     • d. 1,2,2,1
     • e. 2,1,1,2
   • The correct coefficients for the balanced reaction are c. 1,2,1,1, leading us to:
     Pb(NO3)2(aq) + Na2SO4(aq)
     ightarrow PbSO4(s) + 2NaNO3(aq)

Net Ionic Equation

3. Determining the Net Ionic Equation:
   • The net ionic equation focuses on the ions involved in the formation of the precipitate. The choices are:
     • a. $Pb^{4+}(aq) + SO4^{2-}(aq) ightarrow Pb(SO4)_2(s)$
     • b. $Pb^{2+}(aq) + SO4^{2-}(aq) ightarrow PbSO4(s)$
     • c. $Pb^{+}(aq) + SO4^{2-}(aq) ightarrow Pb2(SO4)2(s)$
     • d. $Na^{+}(aq) + NO3^{-}(aq) ightarrow NaNO3(s)$
     • e. None of the above
   • The correct net ionic equation is b. $Pb^{2+}(aq) + SO4^{2-}(aq) ightarrow PbSO4(s)$.

Solubility Comparison

4. Solubility in Water:
   • The solubility of compounds is relevant in predicting which will dissolve in water. The comparison includes:
     • a. AgCl
     • b. Ba3(PO4)_2
     • c. PbCO_3
     • d. CaCl_2
     • e. Fe(OH)_3
   • The compound that is most soluble in water is d. CaCl_2.

Precipitate Mass Calculation

5. Calculating Mass of PbSO4 Precipitate:

   • With the molar mass of $PbSO4$ being 303 g/mol, if unlimited $Na2SO4(aq)$ is combined with 50 mL of 2.5 M $Pb(NO3)_2$, the calculations yield:
     • The number of moles of $Pb^{2+}$ is:
       moles = Molarity imes Volume = 2.5 rac{mol}{L} imes 0.050 L = 0.125 mol
     • Therefore, mass of $PbSO_4$ is:
       mass = moles imes molar mass = 0.125 mol imes 303 g/mol = 37.875 g
   • Round this gives 37.9 g as the correct answer.

6. Making a 2.50 M Solution:

   • To create 100.0 mL of 2.50 M $PbSO_4$, one must determine the grams necessary:
     • molarity = rac{mass (g)}{volume (L) imes molar mass (g/mol)}
     • Rearranging provides mass required:
       mass = molarity imes volume imes molar mass = 2.5 M imes 0.1 L imes 303 g/mol = 75.75 g
   • Thus, the exact requirement is 75.8 g.

Calculating Molarity

7. Finding Molarity from Mass:
   • A solution of 250 mL made with 35.0 g of $PbSO_4$ yields:
     • First calculate moles:
       moles = rac{mass}{molar mass} = rac{35.0 g}{303 g/mol} hickapprox 0.115 mol
     • Then convert to molarity:
       M = rac{0.115 mol}{0.250 L} = 0.46 M
   • Thus, this molarity is 0.46 M.

Dilution Calculation

8. Dilution Calculation for PbSO4:
   • If starting with a 2.5 M solution to achieve 100 mL of a 1.0 M solution, the volume needed can be calculated using:
     • C1V1 = C2V2
     • where $C1 = 2.5 M$, $C2 = 1.0 M$, $V_2 = 100 mL$.
     • Solving for $V1$: V1 = rac{C2V2}{C_1} = rac{1.0 M imes 100 mL}{2.5 M} = 40 mL
   • This indicates 40 mL is required.

Strong Electrolytes

9. Identifying Non-Soluble Electrolyte:
   • Among the following, the identification of a weak electrolyte:
     • a. NaCl
     • b. HNO_3
     • c. CH_3COOH
     • d. LiOH
     • e. K3PO4
   • The substance that is NOT a strong electrolyte is c. CH_3COOH, which is acetic acid, a weak acid that does not fully ionize in solution.

Strong Bases Characteristics

10. Characteristics of Strong Bases:
    • From the options provided, the truth about strong bases is questioned:
      • a. Strong bases are strong electrolytes.
      • b. Strong bases are very concentrated.
      • c. Strong bases are mostly converted to ions when dissolved in water.
      • d. Strong bases will react with strong acids.
      • e. All of the above are true.
    • The false statement about strong bases is b. Strong bases are very concentrated, as concentration can vary and does not define strength.

Neutralization Reaction Calculation

11. Neutralization of NaOH with HCl:
    • To determine the volume of 0.500 M HCl needed to neutralize 50.0 mL of 0.250 M NaOH:
      • The neutralization reaction between NaOH and HCl is a 1:1 ratio.
      • Therefore, moles of NaOH:
        ext{moles NaOH} = 0.250 M imes 0.050 L = 0.0125 mol
      • Thus required moles of HCl is equal to moles of NaOH.
    • To find the volume of HCl needed:
      • M_{HCl} = 0.500 M
      • $volume = rac{moles}{M} = rac{0.0125 mol}{0.500 M} = 0.025 L = 25 mL$
    • Hence, 25.0 mL of HCl is required to complete the neutralization reaction.