Practice Problems 10
Extra-credit will be announced in lab for doing some or all of these questions correctly by March 29 at noon - just before your next test!

    The following questions are designed to help give you a feel for what Keq is all about. It should be noted to you that while most of the reactions in inorganic chemistry don't seem to be reversible, most of those in biochemistry (the chemistry that goes on in our body) are reversible, and thus for those of you interested in the health sciences, this perspective should assist you in your future studies. So learn it now and don't forget it! Anyway, the Keq of a reaction equals the product of all the products' concentrations divided by the product of all the reactants AT THE TIME WHEN EQUILIBRIUM IS THOROUGLY ESTABLISHED.

CHEMICAL REACTIONS PART ONE
[1][2][3]

  1. Concerning Rx [1], This is an interesting type of reaction in which one part of the molecule twists into a different configuation and thereafter can twist back and forth finally establishing an equilibrium mixture of the two forms. Of course if one of those forms is a bit less strained than the other shape, it will be in the majority. Some real-life examples of this include mutarotation in sugars, and various prion diseases such as mad-cow disease. Anyway, pretend you begin with 1 mole of KA and add water up to a liter. You obtain a red solution, which you measure its absorbance at its complementary color and find it to be 0.5 (the dial moves to 0.5). The next morning you find the solution has paled and now the absorbance is 0.25, and after several more readings you find that the readings don't change. Equilibrium has been attained. Write the equation for this reaction's Keq and give the value of the Keq.

  2. Concerning Rx [2], pretend you begin with 1 mole of MN and add water up to a liter. You obtain a blue solution, which you measure its absorbance at its complementary color and find it to be 0.8. The next morning after attaining equilibrium, you find the absorbance to be 0.4.

    1. What is the Keq?

    2. Suppose you added to the above equilibrium 1 mole of N and waited for a new equilibrium to be attained. What would be the Keq?

  3. Concerning Rx [3], which is another "twist" configuration change, pretend you begin with 1 mole of KJ and add water up to a liter. You obtain a purple solution, which you measure its absorbance at its complementary color and find it to be 0.3. What would the final absorbance be if (fill in the blank areas):

    Keq1021½0.1
    Absorbance     

    The following reactions are leading you into the workings of strong and weak acids and bases. What is a strong acid? And what is a weak acid? In what way is sulfuric acid both a strong and a weak acid?

CHEMICAL REACTIONS PART TWO
[4][5][6][7]

  1. SH is a weak acid and when it dissolves in water, the Keq = 0.001. What is the concentration of H+, [H+]? Knowing that pH = -log10[H+], what is the pH of this aqueous solution of SH?

  2. RH is another weak acid, which when dissolved in water causes the pH to be 7. What is the Keq?

  3. Yet another weak acid is TH, which when dissolved in water causes the pH to be 1.0. Does TH dissociate in water more than RH? More that SH? If the pH = 1.0, what is the [H+] of this solution?

  4. A non-fictional weak acid we have used in the course several times is acetic acid, shown at the right, but commonly abbreviated as AcOH or HOAc. In the supermarket, it is vinegar at a 3% solution (quite dilute!). When you dissolve some HOAc into water you obtain a pH = 3.5. What is the [H+]? What is the Keq?

  5. Suppose you added one mole of NaOAc (sodium acetate) to the above solution. Would the Keq change? What would happen to the pH? Would it become more acidic or more alkaline? Would the pH go up or down? What would the pH become?