Chemistry 222. Start mol End mol

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1 Chemistry Spring 019 Exam 3: Chapters 8-10 Name 80 Points Complete problem 1 and four of problems 6. CLEARLY mark the problem you do not want graded. You must show your work to receive credit for problems requiring math. Report your answers with the appropriate number of significant figures. You may ignore activity coefficients in all problems. You MUST complete problem 1. (16 pts.) 1. Find the ph of a solution prepared by dissolving all of the following compounds in water in one beaker and diluting to a volume of L: mol acetic acid (pk a = 4.76), mol sodium acetate, mol HNO 3, and mol KOH. We need to recognize that the strong acid and strong base will react until one is consumed: HNO 3 + NaOH NaNO 3 + H O Start mol End mol Now, the excess strong acid will react: A- + HNO 3 HA + - NO 3 Start mol End mol So, we now have a buffer with mol acetic acid and moles acetate: ph = pk a + log mol A - = log mol = 4.67 mol HA mol In the end, ph =

2 Do four of problems 6. Clearly mark the problem you do not want graded. (16 pts. ea.). Your new employer has asked you to prepare 1.00 L of a ph 1.00 buffer with a total phosphate concentration of M. You have at your disposal only the following compounds Compound K a Molar Mass (g/mol) H 3PO x NaH PO x Na HPO 4 4. x Na 3PO a. Which two compounds would you use to prepare a buffer of ph 1.00 and how many grams of each of the two selected compounds would you need? (1 points) When preparing a buffer, it is best to choose and acid/base pair whose pk a is as close to the target ph as possible. This optimizes buffer capacity. In the case of the items at our disposal, the Na HPO 4/ Na 3PO 4 combination seems best since the pk a for HPO 4 is First, determine the ratio of PO 4 /HPO 4 needed for ph 1.00 by rearranging the Henderson Hasslebach equation.: mol PO 4 = 10 ph-pka = = mol HPO 4 We also know that the total moles of PO 4 and HPO 4 = 1L x M = mol. So: mol PO 4 + mol HPO 4 = and mol PO 4 = mol HPO mol HPO 4 + mol HPO 4 = mol HPO 4 = Solving for each term, we find: mol PO 4 = mol and mol HPO 4 = mol Masses for each salt: mol Na 3PO 4 x g Na 3PO 4 =.46 g Na 3PO 4 1 mol Na 3PO mol Na HPO 4 x g Na HPO 4 = 4.97 g Na HPO 4 1 mol Na HPO 4 b. If you did exactly what you calculated in part (a), you would not get a ph of exactly Why? Explain how you would really prepare this buffer in lab. (4 points) There are two considerations/assumptions we have made that will introduce some uncertainty in our result: (1) we have ignored the contribution of activities and () we have assumed that our initial concentrations of HA and A - are reasonable approximations of the equilibrium concentrations. While these assumptions may be reasonable, they do introduce some error. In practice, we could prepare the buffer as above and add small amounts of strong acid or base to fine tune the ph.

3 3. I've asked you to go into the lab and help me prepare some unknowns for a new acid/base titration experiment we are considering. Unfortunately, I have neglected to label one solution and am nowhere to be found. To identify the solution, you construct the titration curve below by titrating 0.00 ml of the acid solution with standard M NaOH. From the titration curve and the list of possible solution compositions below, identify the composition of the solution. Justify your reasoning by explaining how you were able to rule out each of the imposters and choose the appropriate identity. Solution pk a A: M maleic acid 1.83, 6.07 B: M succinic acid 4.1, 5.64 C: M phosphoric acid.15, 7.0, 1.35 D: M HCl and M acetic acid strong, 4.76 The presence of clearly-resolved breaks in the titration curve indicates the presence of at least two acidic protons with substantially different K a's. Also, the fact that the two equivalence points occur exactly at 0.00 ml increments indicates that the solution contains either a polyprotic acid, or two monoprotic acids present at identical concentrations. This rules out D, since this solution will result in a titration curve with one equivalence point at 6 ml and the second at 0 ml. Solution B can also be ruled out because the relative pk as are close enough that he first equivalence point would be washed out, resulting in a curve showing only a single discrete equivalence point. This leaves solutions A and C, both of which have pk as which are substantially different. How do we rule one out and arrive at an identity? For a polyprotic acid, we would expect the ph halfway to the first equivalence point to equal pk a1 and the ph halfway to the second equivalence point to equal pk a. From the plot, we find these values to be ~. and ~7., respectively corresponding well to solution C and confirming our assignment. The third proton for phosphoric acid is too basic to be titrated with 0.1 M NaOH, so we do not see a third equivalence point. If solution A were the unknown, we would expect the ph values at the two midpoints to be 1.83 and 6.07, which does not match the data. 3

4 4. Calculate the ph of M calcium hydroxide (a strong base). What fraction of the total OH - in this solution is from the calcium hydroxide? You must use charge (or mass) balance to solve this problem! H O = H + + OH -, K w = [H + ][OH - ] Charge Balance: [H + ] + [Ca + ] = [OH - ] [Ca + ] =1.6 x 10-7 M, [OH - ] = K w/[h + ] [H + ] + 3. x 10-7 M = K w [H + ] [H + ] + 3. x 10-7 M [H + ] = K w 0 = [H + ] + 3. x 10-7 M [H + ] - K w Solve for [H + ] using quadratic formula: [H + ] =.87 x 10-8 M, ph = 7.54 If [H + ] =.87 x 10-8 M, the [OH - ] = K w/.87 x 10-8 = 3.49 x 10-7 M. The calcium hydroxide contributes 3. x 10-7 M to this or a fraction of: 3. x 10-7 M = or 91.8% 3.49 x 10-7 M 4

5 5. What is the predominant species present in a solution of sulfurous acid that is buffered at ph 7.00? If the formal concentration of this solution is M, what is the concentration of the predominant species at this ph? For H SO 3, K a1 = 1.3 x 10 - and K a = 6.60 x 10-8 At pk a1 (ph = 1.91), [H SO 3] = [HSO ], while at ph = pk a = 7.17, [HSO ] = [SO 3 ]. Therefore, at an intermediate ph (such as 7.00), HSO 3 - will be the predominant species. We can use an α calculation to figure our its concentration. [HSO ] = (α HSO3)F α HSO3 = [H + ]K a1 [H + ] + [H + ]K a1 + K a α HSO3 = 10-7 (1.3x10 - ) = 0.60 (10-7 ) (1.3x10 - ) + (1.3x10 - )(6.60x10-8 [HSO ] = (α HSO3)F = 0.60 x 0.100M = 0.060M 5

6 6. A weak diprotic acid, H A, has acid dissociation constants of K a1 = and K a = Calculate the ph and molar concentrations of H A, HA, and A at equilibrium for a 0.140F solution of NaHA. First find the [H + ]: Ka 1K af + K w [ ] ( 3.06x10 )( 3.37x10 )( 0.140M) + ( 3.06x10 )( 10 ) 7 H = = = 1.01x10 M K a1 + F 3.06x M ph = -log(1.01x10-7 ) = 6.99 (you could also take the average of the pk as and get the same result) Once we have [H + ], we can find the concentrations by determining the fraction of each species present. [X] = α xf. α HA = + [ H ] + + [ H ] + [ H ] K a1 + K a In all cases, the denominator of the alpha expression is the same: d = [H + ] + K a1[h + ] + K a = (1.01x10-7 ) + ( )( 1.01x10-7 ) + ( )( ) = 3.109x10-11 For H A: α HA = [H + ] = (1.01x10-7 ) = d 3.109x10-11 and [H A] = αhaf = x0.140M = 4.64x10-5 M For HA - : α HA- = K a1[h + ] = ( )( 1.01x10-7 ) = d 3.109x10-11 and [HA - ] = αha-f = 0.999x0.140M = M = M For A : α A = K a = ( )( ) = d 3.109x10-11 and [A ] = αaf = x0.140M = 4.64x10-5 M 6

7 Possibly Useful Information K [ ] af + K H = w K a + ph = ( pk a + pk ) ph pk a + α A Ka1 + F + [ conjugate base ] [ H ] = log αh A = [ weak acid] + + [ H ] + [ H ] Ka1 + K a K a = K ak b = K w + + [ H ] + [ H ] K a1 + K a K w = 1.0 x = [H + ][OH - ] 1 b ± x = 1 a b 4ac a 7

Now, the excess strong base will react: HA + OH - A - + H 2 O Start mol End mol

Now, the excess strong base will react: HA + OH - A - + H 2 O Start mol End mol Chemistry Spring 016 Exam 3: Chapters 8-10 Name 80 Points Complete problem 1 and four of problems -6. CLEARLY mark the problem you do not want graded. You must show your work to receive credit for problems