School of Chemistry, University of KwaZulu-Natal, Howard College Campus, Durban. CHEM191 Tutorial 1: Buffers

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1 School of Chemistry, University of KwaZulu-Natal, Howard College Campus, Durban CHEM191 Tutorial 1: Buffers Preparing a Buffer 1. How many moles of NH 4 Cl must be added to 1.0 L of 0.05 M NH 3 to form a buffer whose ph is 11.00? (Assume that the addition of NH 4 Cl does not change the volume of the solution.) Solution Analyze: Here we are asked to determine the amount of NH + 4 ion required to prepare a buffer of a specific ph. Plan: The major species in the solution will be NH + 4, Cl, and NH 3. Of these, the Cl ion is a spectator (it is the conjugate base of a strong acid). Thus, the NH + 4 NH 3 conjugate acid-base pair will determine the ph of the buffer solution. The equilibrium relationship between NH 4 + and NH 3 is given by the base-dissociation constant for NH 3 : The key to this exercise is to use this K b expression to calculate [NH 4 + ]. Solve: We obtain [OH ] from the given ph: poh = ph = = 3.00 [OH - ] = 1.0 x 10-3 M Because K b is small and the common ion NH + 4 is present, the equilibrium concentration of NH 3 will essentially equal its initial concentration: [NH 3 ] = 0.05 M 1

2 Show working here!!!! [NH 4 ] = M Thus, in order for the solution to have ph = 11.00, [NH 4 + ] must equal M. The number of moles of NH 4 Cl needed to produce this concentration is given by the product of the volume of the solution and its molarity: (1.0L) ( ) = mol 2. Leave out. Calculating the ph of a buffer Acidic Buffers 3. What is the ph of a buffer that is 0.12 M in lactic acid (HC 3 H 5 O 3 ) and 0.10 M in sodium lactate? For lactic acid, K a = Solution The initial and equilibrium concentrations of the species involved in this equilibrium are 2

3 Because K a is small and a common ion is present, we expect x to be small relative to either 0.12 or 0.10 M. Thus, our equation can be simplified to give 4. Suppose you have a solution which is 0.20 M in acetic acid (HAc) and 0.10 M in sodium acetate (NaAc). What will be the ph of this solution? Solution The values of x may often be neglected if they can be shown to be negligibly small. Compare the concentrations to the K a value. In this case the concentrations of the weak acid and its conjugate base are known to the 10-2 place and the K a is to the 10-5 place. Since the difference in magnitude is greater than 100 (actually 3

4 1000 times different), both of this x quantities may be neglected. This simplifies the algebraic expression to: Solving for x gives the hydronium ion concentration: To find the ph, take the negative log the hydronium ion concentration: ph = -log[h 3 O + ] = -log[3.6 x 10-5 ] = 4.44 Alkaline Buffers 5. Suppose you have a alkaline buffer consisting of 0.20 M aqueous ammonia (NH 3 ) and 0.10 M ammonium chloride (NH 4 Cl). What is the ph of the solution? Solution 4

5 Insert the above expressions into the mass action equilibrium expression: The x quantities often may be neglected. Compare the magnitude of the concentrations of NH 3 and NH 4 + to the magnitude of K b. In this case the magnitudes of the concentrations are known to the 10-2 magnitude and the K b is known to the 10-5 magnitude. Since the difference in magnitude is greater than 100 (difference is 1000), the x quantities shown in bold may be neglected. This simplifies the expression to the expression shown: Solve this expression for x which will provide the hydroxide ion concentration: To find the ph, first find the poh by taking the negative log of the hydroxide ion concentration: poh = -log[oh - ] = -log(3.6 x 10-5 ) = 4.44 Next, subtract the poh from to find the ph: ph = poh = =

6 Calculating ph Changes in Buffers 6. A buffer solution is 1.0 M in acetic acid and in sodium acetate. Calculate the change in ph upon adding 0.1 mole of gaseous hydrochloric acid to 1000 ml of this solution. Assume that the volume does not change when the HCl is added. The acid dissociation constant, Ka for CH 3 COOH is 1.8 x [CH 3 COOH] = 1.0M AND [CH 3 COO ] = 1.0M K a = [H + ][CH 3 COO ] =1.8 x 10-5 [CH 3 COOH] [H + ] = = = 1.8 x 10-5 M ph = -log (1.8 x 10-5 ) = 4.74 K a [CH 3 COOH] [CH 3 COO ] (1.8 x 10-5 ) (1.0) (1.0) HCl(aq) H + (aq) + Cl (aq) 0.1 mol 0.1 mol 0.1 mol Originally there were 1.0 mol CH 3 COOH and 1.0 mol CH 3 COO present in 1L of the solution. After neutralization of the HCl acid by CH 3 COO, which we write as CH 3 COO (aq) + H + (aq) CH 3 COOH(aq) 0.1 mol 0.1 mol 0.1 mol The number of moles of acetic acid and the number of the moles of acetate ions present are CH 3 COOH: CH 3 COO _ : ( ) mol = 1.1 mol ( ) mol = 0.9 mol 6

7 Next we calculate the hydrogen ion concentration: [H + ] = = K a [CH 3 COOH] [CH 3 COO ] (1.8 x 10-5 ) (1.1) 0.9 = 2.2 x 10-5 M The ph of the solution becomes Therefore the change in ph = = ph = log (2.2 x 10-5 ) = Calculate the ph of the solution resulting from the addition of ml of mol dm -3 NaOH to ml of mol dm -3 CH 3 COOH. The acid dissociation constant, Ka for CH 3 COOH is 1.76 x n of base = x = n acid = x = mol CH 3 COOH + NaOH CH 3 COO - + H 2 O Start React Final Show full working here!!! ph =

8 8. A buffer solution is 0.20 M in acetic acid and in sodium acetate. a) Calculate the change in ph upon adding 1.0 ml of 0.10 M hydrochloric acid to 10 ml of this solution. The acid dissociation constant, Ka for CH 3 COOH is 1.76 x Initial ph = -(logka) = 4.75 ( show how you arrived at this!) No. of moles of base = dm mol dm -3 = mol = no. moles of acid No. of moles of HCl = dm mol dm -3 = mol CH 3 COO- + H 3 O + CH 3 COOH + H 2 O Start React Final Show full working here!!! ph = 4.71 ph = ph final ph initial = = b) Explain the effect of the addition of the acid in the solution in the question above. The weak base will react with the strong acid according to the following reaction CH 3 COO - + H 3 O + CH 3 COOH + H 2 O The strong acid is converted into the weak acid CH 3 COOH and the ph will decrease. 8

9 9. What would be the ph of a solution prepared by mixing ml of M acetic acid and ml of M sodium acetate? Ka (CH3COOH) = No reaction occurring. After mixing, [ CH 3COO ] x M [ CH 3COOH] x M Show working here!! ph = Calculate the ph of the solution prepared from adding ml of mol dm -3 NaOH to ml of a mol dm -3 CH 3 CH 2 COOH solution (pk a CH 3 CH 2 COOH = 4.88). Answer: Initial moles of CH 3 CH 2 COOH = ml = mol mol 1dm 3 1dm 1000 ml 3 Initial moles of NaOH = ml = mol mol 1dm 3 1dm 1000 ml 3 9

10 Moles CH 3 CH 2 COOH + NaOH CH 3 CH 2 COO - + Na + + H 2 O Initial React Form Equilibrium Therefore after this reaction is complete we have both CH 3 CH 2 COOH and CH 3 CH 2 COO - present. We thus have a buffer. Show working here!!! ph =