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Step by Step Sample Problems

Determining the Mass of Product Formed (use Steps for Solving Stoichiometry Problems)

When aqueous solutions of AgNO3 and KCl are mixed, AgCl precipitates. Calculate the mass of AgCl formed when 1.25 L of 0.500 M AgNO3 and 2.00 L of 0.250 M KCl are mixed.

  1. When the two solutions are mixed together, the resulting solution contains the ions Ag+, NO3, K+, and Cl. KNO3 is soluble and AgCl is not, so therefore AgCl will be the precipitate.

  2. In order to get the net ionic equation I must write out all the ions in part (i).
    Ag+ + NO3 + K+ + Cl –>  AgCl (s) + K+ + NO3
    Ag+ + Cl –>  AgCl (s)

  3. In a 0.500 molar solution of AgNO3, there is 0.500 moles of Ag+.

    1.25 L x 0.500 mol Ag+ = 0.625 mol Ag +
                        L
    In a 0.250 molar solution of KCl, there is 0.250 moles of Cl

    2.00 L x 0.250 mol Cl = 0.500 mol Cl
                                        
     L

  4. Cl will be the limiting reactant because AgCl react in a 1:1 ratio. Since 0.500 mol of Cl is less than 0.625 mol of Ag+, Cl is the limiting reactant.
  5. Using the limiting reactant, we figure out that only 0.500 moles of product will be formed.
  6. 0.500 moles of AgCl x 143.35 g AgCl = 71.7 g AgCl formed
                                  1 mol of AgCl

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Neutralization Reaction

There are six important steps in calculating acid-base reactions. For example, given the problem:   What volume of a 0.200 M HBr solution is needed to neutralize 50.0 mL of 0.500 M KOH?  

  1. Write down the given chemical compounds and then write the reaction that takes place. Remember to balance the chemical reaction.   Compounds that occur in this reaction are HBr and KOH   Balanced equation: HBr(aq)+KOH(aq)–>H2O(l)+KBr(s)   From the balanced equation above, two possible reactions can occur because two products have formed, H2O and KBr. But how do you decide which reaction to use? As you can see   

    K+(aq)+Br(aq)–>KBr(s)

    H+(aq)+OH(aq)–>H2O(l)    You would use the second reaction. Why? According to the solubility rules, KBr is soluble whic means that the reaction won’t occur, but the second reaction does.

  2. Write the net ionic equation by first writing the complete net ionic equation using the balanced reaction. Afterwards, cancel all spectator ions.  

    H+(aq)+Br(aq)+K+(aq)+OH(aq)–>H2O(l)+K+(aq)+Br(aq)

      KBr dissociates completely because it is soluble so K+ and Br- would be the spectator ions. Cancelling these two elements would give the equation   Here is your net ionic equation:  

    H+(aq)+OH(aq)–>H2O(l)

     

  3. Calculate the moles of reactants using not only the net ionic equation but also the volumes and molarities given in the original problem.  

    50.0 mL x 1L/1000 mL x 0.500 mol OH/L KOH= 2.50×10-2 mol OH

     

  4. Determine the limiting reactant if you can. However, most neutralization reactions do not require limiting reactants because they are neutralization reactions.  

    The limiting reactant is not necessary here because this is a neutralization reaction which means the right amount of H+ ions must be added to neutralize the OH ions.

     

  5. Using the net ionic equation and mole ratios, calculate the moles of reactants needed.  

    In the net ionic equation, H+ and OH are in a 1:1 mole ratio. Since there are 2.50×10-2 mol of OH present, and the mole ratio is 1:1, then the mol H+ would also be 2.50×10-2.

     

  6. Finally, using the calculated moles of the reactant, convert it back to grams or volume. However, since this reaction is taking place within a solution, volume would be the best choice.  

    v= 2.50 x 10-2 mol H+/.200 mol H+/L = 1.25 x 10-1 L

     

    So this reaction requires 1.25×10-1L to neutralize 50.0 mL of 0.500 M KOH

     

REMEMBER SIG FIGS MATTER!

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Oxidation-Reduction Reaction

Oxidation-reduction reactions occurring in basic solutions

For example, given the reaction

Cr2O72-(aq)+HNO2(aq)–>Cr3+(aq)+NO3(aq)

1. Split up the reaction into two half reactions. To determine the two half reactions, pair up like elements.

Cr2O72- –> Cr3+

HNO2 –> NO3

2. Before focusing on the hydrogens and oxygens, balance all other elements. Add a two to Cr3+ because Cr2O72- has two Crs and you do not need to add anything to HNO2 or NO3 because both have one nitrogen so it is balanced.

Cr2O72- –> 2Cr3+

HNO2 –> NO3

3. After all other elements are balanced, balance all the oxygens by adding H2O. In these two reactions, we add seven waters to the right side because on the left side there are seven oxygens in dichromate. In the second half reaction, we added a water molecule to HNO2 because NO3 has three oxygens while HNO2 has only two.

Cr2O72- –> 2Cr3+ + 7H2O

H2O + HNO2 –> NO3

4. After all other elements and all oxygens are balanced, balance all the hydrogens by adding H+ ions. We added 14 H+ ions to the left because there are 14 hydrgoens on the right and in the second half reaction we added three H+ ions to the right because there are three hyrdogens on the left.

14H+ + Cr2O72- –> 2Cr3+ + 7H2O

H2O + HNO2–> NO3 + 3H+

5. Balance the charges by adding electrons. If necessary, multiply half reactions by whole numbers to acquire the correct number of charges so that they will cancel out.

6e- + 14H+ + Cr2O72- –> 2Cr3+ + 7H2O

3(H2O + HNO2 –> NO3 + 3H+ + 2e-)

6e- + 14H+ + Cr2O72- –> 2Cr3+ + 7H2O

3H2O + 3HNO2 –> 3NO3 + 9H+ + 6e-

Cr2O72- + 3HNO2 + 5H+ –> 2Cr3+ + 3NO3 + 4H2O

6. Considering that we are trying to achieve half reactions involving basic solutions, we should eliminate the H+ ions by adding OH- ions. Adding the OH- ions will turn the reaction into a basic one.

Cr2O72- + 3HNO2 + 5H+ –> 2Cr3+ + 3NO3 + 4H2O

                       +5OH                             +5OH

Cr2O72- + 3HNO2 + H2O –> 2Cr3+ + 3NO3 + 5OH

7. A very important step, check to see that the elements and charges are balanced.

elements balanced: 2Cr, 14O, 5H, 3N –> 2Cr, 3N, 14O, 5H

charges balanced: (2-)+0+0 –> 2(+3)+3(1-)+5(1-)

                            2- –> 2-

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Neutralization Titration

A student carries out an experiment to standardize a sodium hydroxide solution. To do this, the student weighs out a 1.3009-g sample of potassium hydrogen phthalate (KHC8H4O4, known as KHP.) KHP (molar mass 204.22 g/mol) has one acidic hydrogen. The student dissolves the KHP in distilled water, adds phenolphthalein as in indicator, and titrates the resulting solution with NaOH. The difference between the final and initial buret readings shows that 41.20 mL of the sodium hydroxide solution is required to react exactly with the 1.3009 g KHP. Calculate the concentration of the NaOH solution.

Solution:

Since NaOH is soluble, it splits up into Na+ and OH ions. Also, KHC8H4O4 dissolves in water to form K+ and HC8H4O4 ions. The net ionic equation would be:

HC8H4O4(aq) + OH(aq) g H2O (l) + C8H4O42-(aq)

 Once you find the net ionic equation, you can use stoichiometry to determine the amount of OH ions in the solution. First we would convert the grams of KHC8H4O4 to moles.

1.3009 g KHC8H4O4 x 1 mol KHC8H4O4        = 6.7301×10-3 mols KHC8H4O4

                              204.22 g KHC8H4­O4

Since there are 6.7301×10-3 mols of KHC8­H4O4­, there would also be the same amount for HC8H4O4 because of the 1:1 mole ratio for HC8H4O4 and KHC8H4O4. It also means that 6.7301×10-3 mols of OH must react with the HC8H4O4­­ due to the 1:1 mole ratio for OH and the HC8H4O4­­. One we find the moles of the OH, we can then determine the molarity of the solution.

Molarity of NaOH= mols NaOH  = 6.7301×10-3 mols NaOH = 0.1546 M

                         L solution        4.120×10-2 L

We would need 0.1546M of the NaOH solution to titrate the KHC8H4O4 solution.

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