CHEM 213 Chemical Analysis Practice Exam 2

Transcription

1 CHEM 213 Chemical Analysis Practice Exam (of 10) 2 10 (of 10) 3 10 (of 10) 4 15 (of 15) 5 5 (of 5) 6 10 (of 10) 7 10 (of 10) Σ 70_ KEY Name: (please print)

2 1. Calculate the ph of water containing 0.10 M KCl at 25 ºC. Use Kielland s table of activity coefficients for aqueous solutions on page 9 of this exam for your calculations. (10 points) The relevant equilibrium is: 2H 2 H H - K w K w = a H3+ a H- = [H 3 + ]γ H3+ [H - ]γ H- The autoprotolysis reaction of water tells us that [H 3 + ] = [H - ]. However, the activity coefficients are not equal. The ionic strength of 0.10 M KCl is: µ = ½ [0.1(1) (1) 2 ] = 0.1 M Looking up the values for the activity coefficients of H 3 + and H -, the following equation results: K w = [H 3 + ]γ H3+ [H - ]γ H- = (x)(0.83)(x)(0.76) x = The concentrations of H 3 + and H - are equal and are both greater than M. The activities of H 3 + and H - are not equal in this solution: a H3+ = [H 3 + ]γ H3+ = ( )(0.83) = a H- = [H - ]γ H- = ( )(0.76) = Finally, we calculate the ph: ph = -log a H3+ = -log( ) =

3 2. Consider the following equilibria, in which all ions are aqueous: (1) Ag + + Cl - AgCl(aq) K = (2) AgCl(aq) + Cl - - AgCl 2 K = (3) AgCl(s) Ag + + Cl - K = a. Calculate the numerical value of the equilibrium constant for the reaction AgCl(s) AgCl(aq) (3 points) K total = K 1 K 2 = Ag + + Cl - AgCl(aq) K 1 = AgCl(s) Ag + + Cl - K 2 = AgCl(s) AgCl(aq) K total =? b. Calculate the concentration of AgCl(aq) in equilibrium with excess undissolved solid AgCl. (4 points) The answer to (a) tells us [AgCl(aq)] = c. Find the numerical value of K for the reaction AgCl 2 - AgCl(s) + Cl - (3 points) K total = K 1 K 2 K 3 = AgCl 2 AgCl(aq) + Cl - K 1 = 1/ Ag + + Cl - AgCl(s) K 2 = 1/ AgCl(aq) Ag + + Cl - K 3 = 1/ AgCl 2 AgCl(s) + Cl - K total =? 3

4 3. Sodium hypochlorite (NaCl, the active ingredient of almost all bleaches, was dissolved in a solution buffered to ph Find the ratio [Cl - ]/[HCl] in this solution. (pk a = 7.53 for hypochlorous acid, HCl) (10 points) The pk a of hypochlorous acid, HCl is equal to The ph is known, so the ratio [Cl - ]/[HCl] can be calculated form the Henderson-Hasselbach equation. HCl(aq) + H 2 H 3 + (aq) + Cl - (aq) ph = pk a + log([cl - ]/[HCl]) 6.20 = log([cl - ]/[HCl]) = log([cl - ]/[HCl]) = 10 -log([cl-]/[hcl]) = [Cl - ]/[HCl] = [Cl - ]/[HCl] Finding the ratio [Cl - ]/[HCl] requires knowing only the ph and the pk a. We do not need to know how much NaCl was added, nor the volume. 4

5 4. (a) Find the ph of a solution prepared by dissolving 1.00 g of glycine amide hydrochloride (pk a = 8.00) plus 1.00 g of glycine amide in L. (3 points) H 2 N NH 2 Glycine amide C 2 H 6 N 2 MW = ph = pk a + log([b]/[bh + ] = log[(1.00 g/ g/mol)/(1.00 g/ g/mol)] = 8.37 (b) How many grams of glycine amide should be added to 1.00 g of glycine amide hydrochloride to give 100 ml of solution with ph 8.00? (3 points) ph = pk a + log(mol B/mol BH = log(mol B)/(1.00 g/ g/mol) mol B = =0.423 g glycine amide. (c) What would be the ph if the solution in (a) were mixed with 5.00 ml of M HCl? (3 points) B + H 3 + BH + + H 2 Initial moles Final moles ph = log( / ) = 8.33 (d) What would be the ph if the solution in (c) were mixed with ml of M NaH? (3 points) BH + + H - B + H 2 Initial moles Final moles ph = log( / ) = 8.41 (e) What would be the ph if the solution in (a) were mixed with ml of M NaH? (This is exactly the quantity of NaH required to neutralize the glycine amide hydrochloride.) (3 points) The solution in (a) contains mmol glycien amide hydrochloride and mmol glycine amide. Now we are adding mmol of H -, which will convert all of the glycine amide hydrochloride into glycine amide. The new solution contains = mmol of glycine amide in ml. The 5

6 concentration of glycine amide is ( mmol)/( mmol) = M. The ph is determined by hydrolysis of glycine amide: H 2 N NH 2 + H 2 + H 3 N NH 2 + H x x x x x = K b = K w K a = = 1.58 x 10-6 x = 4.32 x 10-4 M ph = -log(k w /x) = Find the activity (not the activity coefficient) of the (C 3 H 7 ) 4 N + (tetrapropylammonium) ion in a solution containing M (C 3 H 7 ) 4 N + Br - plus M (CH 3 ) 4 N + Cl -. (α x = 800 pm for (C 3 H 7 ) 4 N + ) (5 points) For M (C 3 H 7 ) 4 N + Br - plus M (CH 3 ) 4 N + Cl -, µ = 0.01 M, γ = for an ion of charge ±1 with α = 800 pm. a = (0.005)(0.912) =

7 6. Calculate the ph at each point listed for the titration of ml of M cocaine (K b = ) with M HN 3. The points to calculate are V a = 0.0, 10.0, 20.0, 25.0, 30.0, 40.0, 49.0, 49.9, 50.0, 50.1, and 60.0 ml. (10 points) N CH 3 H CH 3 + H K B = 2.6x H - N The titration reaction is B + H + BH + Representative calculations: At V a = 0.0 ml: B + H 2 BH + + H x x x x 2 /(0.100 x) x = , ph = -log(k w /x) = At V a = 20.0 ml: B + H + BH + Initial: Final: ph = pk a (for BH + ) + log([b]/[bh + ]) ph = log(30.0/20.0) = 8.59 At V a = V c = 50 ml: All B has been converte to the conjugate acid BH +. The formal concentration of BH + is (100/150)(0.100) = M. The ph is determined by the reaction BH + + H 2 B + H x x x x 2 /( x) = K a = K w /K b x = ph = At V a = 51.0 L, there is excess H 3 + : [H 3 + ] = (1.0,151.0)(0.200) = , ph = 2.88 V a (ml) ph V a (ml) ph V a (ml) ph

8 7. How many milliliters of M KH should be added to 3.38 g of oxalic acid to give a ph of 4.40 when diluted to 500 ml? (10 points) H H xalic acid (H 2 x) MW = pk 1 = pk 2 = The desired ph is above pk 2. We know that a 1:1 mole ratio of Hx - :x 2- would have ph = pk 2 = If the ph is to be 4.40, there must be more x 2- than Hx - present. We must add enough base to convert all of the H 2 x to Hx -, plus enough additional base to convert the right amount of Hx-0- to x 2-. H 2 x + H - Hx - + H 2 Hx - + H - x 2- + H 2 In 3.38 g of H 2 x there are mol. The volume of M KH needed to react with this much H 2 x to make Hx - is: ( mol)(0.800 mol/l) = ml. To produce a ph of 4.40 requires: Hx - + H - x 2- Initial moles: x 0 Final moles: x 0 x ph = pk 2 = log([x 2- ]/[Hx - ]) 4.40 = log(x/( x) x = mol The volume of KH needed to deliver mole is: ( mol)/( M) = ml. The total volume of KH needed to bring the ph up to 4.40 is = ml 8

9 9