School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus. CHEM120R Tutorial 2: Buffers and Solubility

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1 School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus CHEM120R Tutorial 2: Buffers and Solubility Buffers 1. What would be the ph of a buffer solution containing M NH + and M NH? Ka (NH + ) = ph [NH ] pka log [NH ] What would be the ph of a solution prepared by mixing ml of M acetic acid and 25.0 ml of M sodium acetate? Ka(CH COOH) = No reaction occurring. After mixing, 25.0 [ CHCOO ] x M [ CHCOOH] x M ph pka log

2 . You have ml of a buffer solution of ph = 5.00 with [CH COOH] = M and [CH COONa] = M. What would be the ph after the addition of 10.0 ml of 0.05 M NaOH? CH COOH + NaOH CH COONa + H 2 O I C E Final ph log 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 COOH is 1.8 x [CH COOH] = 1.0M AND [CH COO ] = 1.0M K a = [H + ][CH COO ] =1.8 x 10-5 [CH COOH] [H + ] = = = 1.8 x 10-5 M ph = -log (1.8 x 10-5 ) =.7 K a [CH COOH] [CH COO ] (1.8 x 10-5 ) (1.0) (1.0) 2

3 HCl(aq) H + (aq) + Cl (aq) 0.1 mol 0.1 mol 0.1 mol Originally there were 1.0 mol CH COOH and 1.0 mol CH COO present in 1L of the solution. After neutralization of the HCl acid by CH COO, which we write as CH COO (aq) + H + (aq) CH 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 COOH: CH COO _ : ( ) mol = 1.1 mol ( ) mol = 0.9 mol Next we calculate the hydrogen ion concentration: [H + ] = = K a [CH COOH] [CH 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 = = 0.08 ph = log (2.2 x 10-5 ) =.66

4 Solubility (Please refer to a table for the solubility product constants) 5. Calculate the molar solubility of silver bromide in water at 20 C. The solubility product at this temperature is Silver bromide dissolves and dissociates according to the reaction For which AgBr(s) Ag + + Br - K sp = [Ag + ] [Br - ] The solubility, or concentration of dissolved silver bromide, is equal to the concentration of free silver ion: Solubility = [AgBr] dissolved = [Ag + ] According to the reaction stoichiometry, silver ion and bromide ion are formed in equal quantities. Since there is no other source of these ions and since they do not undergo reactions with any other substances, their concentrations are equal. [Br - ] = [Ag + ] Substituting in the equilibrium expression gives K sp = [Ag + ] 2 [Ag + ] = solubility = K sp 5.0 x x10 7 M 6. The solubility of lead (II) chloride (PbCl 2) is PbCl 2? The dissolution-precipitation equilibrium is given by M. What is the K sp of For which PbCl 2 (s) Pb Cl ˉ K sp = [Pb 2+ ] [Cl ˉ] 2

5 From the reaction stoichiometry, [Cl ˉ ] = 2[Pb 2+ ] Substituting in the equilibrium expression gives K sp = [Pb 2+ ] (2[Pb 2+ ]) 2 = [Pb 2+ ] = (0.016) = What must be the concentration of Ag + to just start precipitation of AgCl in a 1.0x10 - M solution. AgCl just starts to precipitate when the ion product just exceeds K sp K sp = [Ag + ][Cl - ] 1.8 x10-10 = [Ag + ] [Ag + ] = M 8. Calculate the solubility, in g/100 ml, of calcium iodate in water at 20 C. Calcium iodate dissolves and dissociates according to the reaction Ca(IO ) 2 (s) Ca IO ˉ Solubility Equilibria for which K sp = [Ca 2 ] [IO ˉ] 2 The solubility, or concentration of dissolved calcium iodate, is equal to the concentration of calcium ion: Solubility = [Ca(IO ) 2 ] dissolved = [Ca 2+ ] According to the reaction stoichiometry, twice as much iodate as calcium is formed on dissolution. Thus [IO ˉ] = 2[Ca 2+ ] Substituting in the equilibrium expression gives 5

6 K sp = [Ca 2+ ](2[Ca 2+ ]) 2 = [Ca 2+ ] [Ca 2+ ] = solubility = 7 Ksp 7.1 x x 10 M Converting to the desired concentration units, we obtain Solubility = ( mol/l)(89.9g/mol) = 2.2g/L = 0.22g/100mL 9. Calculate the molar solubility of lead iodide in (a) water and (b) in M sodium iodide solution. The K sp for PbI 2 is (a) The dissolution-precipitation equilibrium is given by PbI 2 (s) Pb I ˉ For which K sp = [Pb 2+ ] [I ˉ] 2 The solubility, or concentration of dissolved PbI 2, is equal to the concentration of Pb 2+ or one-half the concentration of I ˉ ): solubility = [PbI 2 ] dissolved = [Pb 2+ ] From the reaction stoichiometry, [I ˉ ] = 2[Pb 2+ ] Substituting in the equilibrium expression gives K sp = [Pb 2+ ] (2[Pb 2+ ]) 2 = [Pb 2+ ] [Pb 2+ ] = solubility = 9 K sp 7.9 x M (b) The same equilibrium expression holds: K sp = [Pb 2+ ] [I ˉ ] 2 There are now two sources of iodide: the NaI and the PbI 2. The amount of iodide coming from PbI 2 is small compared to that from the NaI. Thus [I ˉ] = C nal + 2[Pb 2+ ] C nal = M 6

7 Then 9 K sp [Pb x 10 ] = solubility = M [I ] (0.200) Note that the solubility has decreased markedly (four orders of magnitude) upon the addition of an excess of Iˉ. 10. The molar solubility of silver sulfate is M. Calculate the solubility product of this salt. Ag 2 SO (s) 2Ag SO K sp = [Ag + ] 2 [SO 2- ] [Ag + ] = M = M SO 2- = M Substituting in the equilibrium expression gives K sp = [ ] 2 ( )= Calculate the solubility of Zn(OH) 2 (K sp = ) in pure water. Zn(OH) 2 Zn OH - [Zn 2+ ] [OH - ] 2 = as [OH - ] = 2 [Zn 2+ ] substituting in above equation we get [Zn 2+ ] (2 [Zn 2+ ]) 2 = [Zn 2+ ] = / = [Zn(OH) 2 ] = [Zn 2+ ] = ( ) 1/ = mol dm If 200 ml of 0.00 M BaCl 2 are added to 600 ml of M K 2 SO, will precipitation occur? (Ksp = ). Know your rules: In a solution, (Q = reaction quotient-ion product) If Q = K sp, the system is at equilibrium and the solution is saturated 7

8 ( no precipitation). If Q < K sp, more solid will dissolve until Q = K sp (unsaturated no precipitation clear solution) If Q > K sp, the salt will precipitate until Q = K sp supersaturated solution- ppt will occur. The net ionic equation is: Ba 2+( aq) + SO 2- (aq) BaSO (s) No. of moles of Ba 2+ = 0.00 mol dm dm = mol No. of moles of SO 2- = (0.008 mol dm - ) (0.600 dm ) = mol [Ba 2+ ] = mol/0.800 dm = mol dm - [SO 2- ] = mol/0.800 dm = mol dm - Compare Q and Ksp BaSO (s) Ba 2+( aq) + SO 2- (aq) Q = ( )( ) = Ksp = Q > Ksp, BaSO will precipitate out of solution until [Ba 2+ ][SO 2-] = Additional problems 1. Calculate the ph of the solution prepared from adding ml of mol dm - NaOH to ml of a mol dm - CH CH 2 COOH solution (pk a CH CH 2 COOH =.88). Answer: Initial moles of CH CH 2 COOH = ml = mol dm 1dm 1000 ml Initial moles of NaOH = ml dm = mol 1dm 1000 ml 8

9 Moles CH CH 2 COOH + NaOH CH CH 2 COO - + H 2 O Initial React Form Equilibrium Therefore after this reaction is complete we have both CH CH 2 COOH and CH CH 2 COO - present. We thus have a buffer. Using the Henderson-Hasselbalch equation. ph =.88 + log = mol/v mol/v 2. Calculate the molar solubility of PbCl 2 given the relevant K sp value is Answer: We are being asked to calculate [PbCl 2 ]. PbCl 2 (s) Pb 2+ (aq) + 2Cl - Therefore K sp = [Pb 2+ ] [Cl - ] 2 [Cl - ] = 2 [Pb 2+ ], substitute into the K sp equation K sp = [Pb 2+ ](2 [Pb 2+ ]) 2 = [Pb 2+ ] 9

10 [Pb 2+ ] = [Pb 2+ ] = = mol dm - [PbCl 2 ] = mol dm -. Calculate the ph at the following points in the titration of ml of M HNO with M NaOH: (a) before the addition of any NaOH, [H O + ] = [HNO ] = mol dm - ph = -log(0.100) = 1.00 (b) after the addition of ml of M NaOH and Initial no. of mol of H O + = (0.100 mol dm - )(0.025 dm ) = mol No. of mol of OH- added = (0.100 mol dm - )(0.010 dm ) = mol No. of mol of H O + remaining = = mol [H O + ] = mol/0.05 dm = mol dm - ph = -log(0.0286) = 1.7 (c) after the addition of ml of M NaOH. Initial no. of mol of H O + = (0.100 mol dm -) (0.025 dm ) = mol No. of mol of OH - added = (0.100 mol dm - )( dm ) = mol No. of mol of OH - in excess = = mol [OH - ] = mol/ dm = mol dm - poh = -log( ) =.0 ph =