Worksheet 6. For the dissociation of the second proton, the following equilibrium must be considered:
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1 1. A 7.0 mass % solution of H 3 PO 4 in water has a density of g / ml. Calculate the ph and the concentrations of all species present (H 3 PO 4, H 2 PO 4 -, HPO 4 2-, PO 4 3-, H 3 O +, and OH - ) in the solution. You will need to look up the equilibrium constants. H 3 PO 4, 98.0 amu Assume L of solution Mass of solution = L 1000 ml / 1 L g / ml = g Mass of H 3 PO 4 = (0.070)( g) = g H 3 PO 4 mol H 3 PO 4 = g H 3 PO 4 1 mol H 3 PO 4 / g H 3 PO 4 = mol H 3 PO 4 [H 3 PO 4 ] = mol H 3 PO 4 / L = M For the dissociation of the first proton, the following equilibrium must be considered: H 3 PO 4 (aq) + H 2 O(l) H 3 O + (aq) + H 2 PO - 4 (aq) initial (M) equil (M) x x x K a1 = [H 3 O + ][H 2 PO 4 - ] / [H 3 PO 4 ] = = x 2 / x x 2 + ( )x - ( ) = 0 Solve for x using the quadratic formula. x = and , only the positive value of x has a physical meaning x = M = [H 2 PO 4 - ] = [H 3 O + ] For the dissociation of the second proton, the following equilibrium must be considered: H 2 PO - 4 (aq) + H 2 O(l) H 3 O + (aq) + HPO 2-4 (aq) initial (M) change (M) -y +y +y equil (M) y y y K a2 = [H 3 O + ][HPO 4 2- ] / [H 2 PO 4 - ] = = ( y)(y) / y (0.0708)(y) / = y y = M = [HPO 4 2- ] For the dissociation of the third proton, the following equilibrium must be considered: HPO 2-4 (aq) + H 2 O(l) H 3 O + (aq) + PO 3-4 (aq) initial (M) change (M) -z +z +z equil (M) ( ) - z z z K a3 = [H 3 O + ][PO 4 3- ] / [HPO 4 2- ] = = ( z)(z) / ( ) - z (0.0708)(z) / z = M = [PO 4 3- ] [H 3 PO 4 ] = x = = 0.67 M 1
2 [H 2 PO 4 - ] = [H 3 O + ] = M = M [HPO 4 2- ] = M [PO 4 3- ] = M [OH - ] = K w / [H 3 O + ] = / = M ph = -log[h 3 O + ] = -log(0.0708) = Neutralization reactions involving either a strong acid or a strong base go essentially to completion, and therefore we must take such neutralizations into account before calculating concentrations in mixtures of acids and bases. Consider a mixture of 3.28 g of Na 3 PO 4 and ml of M HCl. Write balanced net ionic equations for the neutralization reactions, and calculate the ph of the solution. Na 3 PO 4, amu 3.28 g Na 3 PO 4 1 mol Na 3 PO 4 / g Na 3 PO 4 = mol = 20.0 mmol Na 3 PO ml mmol / ml = 54.0 mmol HCl H 3 O + (aq) + PO 3-4 (aq) HPO 2-4 (aq) + H 2 O(l) before (mmol) change (mmol) after (mmol) H 3 O + (aq) + HPO 2-4 (aq) H 2 PO - 4 (aq) + H 2 O(l) before (mmol) change (mmol) after (mmol) H 3 O + (aq) + H 2 PO - 4 (aq) H 3 PO 4 (aq) + H 2 O(l) before (mmol) change (mmol) after (mmol) [H 3 PO 4 ] = 14.0 mmol / ml = 0.47 M; [H 2 PO 4 - ] = 6.0 mmol / ml = M H 3 PO 4 (aq) + H 2 O(l) H 3 O + (aq) + H 2 PO - 4 (aq) initial (M) equil (M) x x x K a = [H 3 O + ][H 2 PO 4 2- ] / [H 3 PO 4 ] = = x( x) / ( x) x x - ( ) = 0 Solve using the quadratic formula. x = ph = -log[h 3 O + ] = -log( ) =
3 3. A L sample of HF gas at 20.0 C and atm pressure was dissolved in enough water to make 50.0 ml of hydrofluoric acid. What is the ph of the solution? PV = nrt, n = PV / RT = (0.601 atm)(1.000 L) / ( L atm K -1 mol -1 )(293.1 K) n = mol HF 50.0 ml 1.00 L / 1000 ml = L [HF] = mol HF / L = M HF(aq) + H 2 O(l) H 3 O + (aq) + F - (aq) initial (M) equil (M) x x x K a = [H 3 O + ][F - ] / [HF] = = x 2 / x x 2 + ( )x - ( ) = 0 Solve for x using the quadratic formula. x = ph = -log[h 3 O + ] = -log(0.0131) = = Ethylenediamine (NH 2 CH 2 CH 2 NH 2, abbreviated en) is an organic base that can accept protons: en(aq) + H 2 O(l) enh + (aq) + OH - (aq) K b1 = enh + (aq) + H 2 O(l) enh 2 2+ (aq) + OH - (aq) K b2 = Consider the titration of 30.0 ml of M ethylenediamine with M HCl. Calculate the ph after additions of the following volumes of acid, and construct a qualitative plot of ph versus milliliters of HCl added. (a) 0.0 ml en(aq) + H 2 O(l) enh + (aq) + OH - (aq) initial (M) equil (M) x x x K b = [enh + ][OH - ] / [en] = = (x)(x) / x x 2 + ( )x - ( ) = 0 Use the quadratic formula to solve for x. x = [OH - ] = x = M [H 3 O + ] = K w / [OH - ] = / = M ph = -log[h 3 O + ] = -log( ) =
4 (b) 15.0 ml (30.0 ml)(0.100 mmol / ml) = 3.00 mmol en (15.0 ml)(0.100 mmol / ml) = 1.50 mmol en Halfway to the first equivalence point, [OH - ] = K b1 [H 3 O + ] = K w / [OH - ] = / = ph = -log[h 3 O + ] = -log( ) = (c) 30.0 ml At the first equivalence point ph = pk a1 + pk a2 / 2 = 9.14 (d) 45.0 ml Halfway to the first and second equivalence points, [OH - ] = K b2 = M [H 3 O + ] = K w / [OH - ] = / = M ph = -log[h 3 O + ] = -log( ) = 7.57 (e) 60.0 ml At the second equivalence point only the acidic enh 2 Cl 2 is in solution. For enh 2 2+, K a = K w / K b for enh + = K w / K b2 = / = [enh 2 2+ ] = 3.00 mmol / (30.0 ml ml) = M enh 2+ 2 (aq) + H 2 O(l) H 3 O + (aq) + enh + (aq) initial (M) equil (M) x x x K a = [H 3 O + ][enh + ] / [enh 2+ 2 ] = = (x)(x) / x x 2 / Solve for x. x = [H 3 O + ] = ( )(0.0333) = M ph = - log[h 3 O + ] = -log( ) = 4.52 (f) 75.0 ml excess HCl (75.0 ml ml)(0.100 mmol / ml) = 1.50 mmol HCl = 1.50 mmol H 3 O + [H 3 O + ] = 1.50 mmol / (30.0 ml ml) = M ph = -log[h 3 O + ] = -log(0.0143) = A 40.0 ml sample of a mixture of HCl and H 3 PO 4 was titrated with M NaOH. The first equivalence point was reached after 88.0 ml of base, and the second equivalence point was reached after ml of base. 4
5 (a) What is the concentration of H 3 O + at the first equivalence point? The first equivalence point is reached when all the H 3 O + from HCl and H 3 O + from the first ionization of H 3 PO 4 is consumed. At the first equivalence point ph = pk a1 + pk a2 / 2 = 4.66 [H 3 O + ] = 10 -ph = 10 (-4.66) = M (88.0 ml)(0.100 mmol / ml) = 8.80 mmol NaOH are used to get to the first equivalence point. (b) What are the initial concentrations of HCl and H 3 PO 4 in the mixture? mmol (HCl + H 3 PO 4 ) = mmol NaOH = 8.8 mmol mmol H 3 PO 4 = (126.4 ml ml)(0.100 mmol / ml) = 3.84 mmol mmol HCl = ( ) = 4.96 mmol [HCl] = 4.96 mmol / 40.0 ml = M [H 3 PO 4 ] = 3.84 mmol / 40.0 ml = M (c) What percent of the HCl is neutralized at the first equivalence point? 100% of the HCl is neutralized at the first equivalence point. (d) What is the ph of the mixture before addition of any base? H 3 PO 4 (aq) + H 2 O(l) H 3 O + (aq) + H 2 PO - 4 (aq) initial (M) equil (M) x x x K a1 = [H 3 O + ][H 2 PO 4 - ] / [H 3 PO 4 ] = = ( x)(x) / x x x - ( ) = 0 Use the quadratic formula to solve for x. x = [H 3 O + ] = x = = M ph = -log[h 3 O + ] = -log(0.129) = 0.89 (e) Sketch the ph titration curve, and label the buffer regions and equivalence points. (f) What indicators would you select to signal the equivalence points? Bromcresol green or methyl orange are suitable indicators for the first equivalence point. Thymolphthalein is a suitable indicator for the second equivalence point. 6. In qualitative analysis, Ca 2+ and Ba 2+ are separated from Na +, K +, and Mg 2+ by adding aqueous (NH 4 ) 2 CO 3 to a solution that also contains aqueous NH 3. Assume that the concentrations after mixing are M (NH 4 ) 2 CO 3 and 0.16 M NH 3. 5
6 (a) List all the Bronsted-Lowry acids and bases present initially, and identify the principal reaction. HCO 3 - (aq) + OH - (aq) CO 3 2- (aq) + H 2 O(l) (b) Calculate the ph and the concentrations of all species present in the solution. mol HCO 3 - = (0.560 mol / L)( L) = mol HCO 3 - mol OH - = (0.400 mol / L)( L) = mol OH - HCO - 3 (aq) + OH - (aq) CO 2-3 (aq) + H 2 O(l) before rxn (mol) change (mol) after rxn (mol) mol HCO 3 - = = mol [HCO 3 - ] = mol / L = M [CO 3 2- ] = mol / L = M HCO - 3 (aq) + H 2 O(l) H 3 O + (aq) + CO 2-3 (aq) initial (M) equil (M) x x x K a = [H 3 O + ][CO 3 2- ] / HCO 3 - ] = = x( x) / x x(0.200) / Solve for x. x = [H 3 O + ] = M ph = -log[h 3 O + ] = -log( ) =
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