Key CH3NH2(aq) H2O(l) CH3NH3 (aq) OH - (aq) Kb = 4.38 x 10-4 In aqueous solution of methylamine at 25 C, the hydroxide ion concentration is 1.50 x 10-3 M. In answering the following, assume that temperature is constant and that volumes are additive. (a) Write the equilibrium constant expression for the reaction above. K b = CH 3NH 3 [OH ] [CH 3 NH 2 ] (b) Determine the initial concentration of methylamine. Since the solution is a weak base, construct an ICE table [CH 3 NH 2 ] CH 3 NH 3 [OH ] Initial [CH 3 NH 2 ] i 0 0 Change x x x Equilibrium [CH 3 NH 2 ] i x x x Substitute the expression of the equilibrium concentrations for each chemical species into the equilibrium constant expression for K b. K b = CH 3NH 3 OH CH 3 NH 2 = x (x) = 4.38 10 4 [CH 3 NH 2 ] i x Since we are looking for the initial concentration of methylamine, [CH 3 NH 2 ] i, the value of x must be known. In the problem, the [OH ] is given as 1.50 10 3 M. This is equal to x. Hence, we can now calculate [CH 3 NH 2 ] i from the equation. Solving the equation, 1.50 10 3 2 = 4.38 10 4 [CH 3 NH 2 ] i 1.50 10 3 [CH 3 NH 2 ] i = 6. 60 10 3 M
(c) Determine the percent ionization of methylamine in the above solution. Percent ionization is the percentage of the initial concentration of the base that dissociates to form CH 3 NH 3 and OH given by the following formula: % ionization = amount dissociated initial amount 100% = x [CH 3 NH 2 ] i 100% = 1.50 10 3 100% 6.60 10 3 % ionization = 22. 7% (d) What is the number of moles of HCl that should be added to 100. ml of 0.100 M CH3NH2 to produce a solution buffered at ph = 10.125. A base buffer solution must contain both the weak base (CH3NH2) and its conjugate acid (CH3NH3 ). Adding HCl into the weak base solution will produce the conjugate acid as shown by the following net ionic equation. CH 3 NH 2 H CH 3 NH 3 The amount of HCl added must not exceed the initial concentration of the CH3NH2, otherwise, no buffer will be formed. The ratio of the weak base and its conjugate acid will determine the ph. In other words, the problem basically asks for the amount of HCl that must be added so that the ratio is just right to give a ph=10.125. Calculating ph of buffer solutions is easier with the use of the Henderson-Hasselbach equation: ph = pk a log [base] [acid] = pk a log [CH 3NH 2 ] [CH 3 NH 3 ] pk a = log K a. Since K a is not given, the value can be calculated using the following equation: K a = Kw where Kw = 1 10 14 while K K b = 4.38 10 4. Therefore b pk a = 10.641
If we let x = moles HCl added, the stoichiometry of the reaction is shown by the table below: Initial moles CH 3 NH 2 moles HCl moles CH 3 NH 3 (0.100 L)(0.100 M) =0.0100 moles x 0 Change x x -x Final 0.0100 x 0 completely used up (Limiting Reactant) x Remember that in the log ratio, the moles of the species can be used instead of a concentration because the volume units will cancel out anyway. ph = pk a log [CH 3NH 2 ] [CH 3 NH 3 ] 0.0100 x 10.125 = 10.641 log x x = 0. 00766 moles HCl
(e) 125.0 ml of 0.200 M CH3NH3Cl is titrated with 0.100 M NaOH to the equivalence point and beyond the equivalence point. We have to identify the acid-base reaction involved in this particular titration. Notice that the acid in the problem is a salt. CH 3 NH 3 is the one that will react with theoh, Na and Cl - are weaker conjugates. CH 3 NH 3 OH CH 3 NH 2 H 2 O (i) Find the ph of a 0.200 M CH3NH3Cl solution. Initially the flask only contains the acid salt solution (CH 3 NH 3 ). Hence, the ph can be solved from the dissociation of the acid salt in H2O using the ICE table below. CH 3 NH 3 H 2 O CH 3 NH 2 H 3 O CH 3 NH 3 [CH 3 NH 2 ] H 3 O Initial 0.200 M 0 0 Change x x x Equilibrium 0.200 x x x K a = K w K b = x 2 0.200 x = 1 10 14 4.38 10 4 x = 2.14 10 6 M ph = log[ H 3 O ] = log 2.14 10 6 = 5. 67
(ii) Find the volume of NaOH needed to reach the equivalence point. CH 3 NH 3 OH CH 3 NH 2 H 2 O At the equivalence point, there is no limiting or excess reactant, both the CH 3 NH 3 and OH completely used up in the reaction. Hence, to calculate the volume needed to reach equivalent point, the following equation can be used because the ratio of acid to base is 1:1 moles CH 3 NH 3 = moles OH (C V) CH3 NH 3 = (C V) OH 0.200 M 125.0 ml = 0.100 M V OH V OH = 250. 0 ml (iii) Find the ph at the equivalence point. At equivalence point, the major species that would affect the ph is only CH 3 NH 2. Since this is a weak base and can dissociate in water, the ph is expected to be greater than 7. Since the moles of CH 3 NH 3 that reacted is the same number of moles of CH 3 NH 2 produced, moles CH 3 NH 3 = 0.200 M 0.125 L = 0.0250 moles moles CH 3 NH 2 = 0.0250 moles CH 3 NH 2 H 2 O CH 3 NH 3 OH The ph can be calculated from the ICE table of the dissociation of the weak base. The initial concentration of CH 3 NH 2 must be calculated based on the number of moles produced and the total volume (original volume of the acid salt volume of the base added) of the solution at this point. [CH 3 NH 2 ] i = 0.0250 moles = 0.0667 M 0.125 L 0.250 L
[CH 3 NH 2 ] CH 3 NH 3 [OH ] Initial 0.0667 0 0 Change x x x Equilibrium 0.0667 x x x K b = x 2 = 4.38 10 4 0.0667 x x = 0.00541 M poh = log 0.00541 = 2.27 ph = 11.73 (iv) Sketch a rough curve for the above titration on the graph below and indicate the initial ph, the ph at half-way to the equivalence point and the ph at the equivalence point. A titration curve can be sketched based on four points only. The initial point The volume of NaOH added is zero. The initial ph is calculated from part (e)(i), ph = 5.67 The equivalence point The volume of NaOH added is 250.0 ml based on calculation in part (e)(ii) The ph at equivalence point is calculated from part (e)(iii), ph=11.73 The half-equivalence point The volume is half of the volume at equivalence point, 125.0 ml Since the ratio of the base and acid is equal to 1 at this point, the Henderson- Hasselbach equation reduces to ph = pka Hence, the ph at half-equivalence point is 10.64 Beyond equivalence Assume a volume greater than the equivalence volume. The difference between the assumed volume and the equivalence volume is the excess NaOH unreacted in the solution. This will determine the ph of the solution. Let s say, we have added 400.0 ml. The excess NaOH would then be 400.0 250.0 ml. Hence the excess base would be 150.0 ml.
The concentration of the base must then be recalculated considering the total volume of the solution at this point. OH = moles excess NaOH total volume of solution = 0.150 L (0.100 M) = 0.0286 M (0.125 L 0.400 L) poh = log 0.0286 = 1.54 ph = 14. 00 1. 54 = 12. 46