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Now that that is out of the way, we are ready to begin…. Start praying now!. 17.1 Common-Ion Effect- sample 17.2 17.2 Buffered solutions resist change in pH Need: #1 Weak acid or weak base #2 Salt #3 Common ion #1 & #2. Buffers. Ex. NH 4 Cl + NH 4 OH
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Now that that is out of the way, we are ready to begin…. Start praying now! • 17.1 Common-Ion Effect- sample 17.2 • 17.2 Buffered solutions resist change in pH • Need: • #1 Weak acid or weak base • #2 Salt • #3 Common ion #1 & #2
Buffers • Ex. NH4Cl + NH4OH NaCl + HCl HF + KF NaC2H3O2 + HC2H3O2
How it works • HX H+1 + X-1 • Add H+ shift to the left: H++X- HX • Add OH- shift to right OH- + HX H2O + X-
Buffer problem • What is the pH of a mixture of 0.040M HF and 0.70M KF solution? (Ka = 3.53 x 10-4) • HF H+1 + F- KF K+ + F- • 0.040 x x 0.70 0.70 0.70 • Ka = [H+1][F-] = (x)(0.70 + x) =3.53 x 10-4 • [HF] (0.040 – x ) • 0.70x = 1.41 x 10-5 • x = 2.02 X 10-5 pH = -log(2.02 X 10-5) • pH = 4.69 (4.70 w/ quad) +x -x
Buffer Capacity • Depends on concentration HC3H5O3 H+ + C3H5O3- .12 M x x NaC3H5O3 Na+ + C3H5O3- .10 M .10 .10 Ka = [H+][C3H5O3-1] = (x)(.10+x) = 1.4x10-4 [HC3H5O3] (.12 – x ) 1 x= 1.7x 10-4 pH= 3.77
OR Use the Henderson- Hasselbalch Hasselhoff equation log [CB] [Acid] • pH= pKa + log [Base] [Acid] =-log(1.4x 10-4) + log (.10/.12) = 3.85 + (-.08) = 3.77 log [A-] [HA]
Sample 17.4 Kb = [NH4+][OH-] pH= 9.00 [NH3] pOH= 5.00 (OH-)= 1x10-5M 1.8x 10-5= (x)(1.0x 10-5) .10M X= .18 M/L * 2.0 L = .36 mol NH4Cl
Addition of SA. Or SB. To Buffers HC2H3O2 H+ + C2H3O2-1 HAc H+ + Ac-1 .3 x x NaAc Na+ + Ac- .3 .3 .3 HC2H3O2 + OH-1 H2O + C2H3O2-1 Int. .300 .02 .300 Rxn. -.02 -.02 +.02 End .28M 0 .32 M Ka= (x)(.28) = 1.8x 10-5 x= 2.1x10-5 (.32) pH= 4.68
17.3 Titration Calculations • Titration
Titration calc. SA vs SB • Given: 50.0ml of 0.100M HCl. • 1.) What is the pH of this solution before any base is added? • HCl H+ + Cl- • 0.100 0.100 0.100 • pH = -log (0.100) • = 1.00
2.) What is the pH after 25.0ml of 0.100MNaOH has been added? • 50.0ml x 1 liter x 0.100 mol = 0.005 mol H+ • 1000ml 1 liter • 25.0ml x 1 liter x 0.100 mol = 0.0025 mol OH- • 1000ml 1 liter • H+ +OH- H2O start .005mol .0025mol 0
∆ -.0025mol -.0025mol +.0025mol • End .0025 0 .0025 • pH = -log [ H+] • = -log(.0025/.075liters) • = 1.48 • 3.) ….pH after 49.9ml of 0.100M NaOH has been added? • H+ +OH- H2O • I .005mol .00499 0
∆ -.00499 -.00499 +.00499 • End .00001 mol 0 +.00499 • pH = -log (.00001mol/.0999liter) • = 4.00 • 4.) …after 50.0ml of 0.100M of NaOH? • pH = 7.00 • Mol of acid = mol of base
5.) …after 50.1ml of 0.100MNaOH have been added? • H+ +OH- H2O • I .00500mol .00501 0 • ∆ -.00500 -.00500 +.00500mol • E 0 .00001mol +.00500 XS is BASE pOH = -log (.00001mol/.1001liter) = 4.00 pH = 10.00
6.)…after 75.0ml of NaOH have been added? • pH = 12.30
WEAK ACID vs STRONG BASE • 25.0ml of 0.100M HC2H3O2 (Ka = 1.8 x 10-5) is being titrated with 0.100M NaOH. • 1.) What is the pH before any NaOH has been added? (pH of a weak acid) • HC2H3O2 H+ + C2H3O2- • 0.100 x x • x2= Ka = 1.8 x 10-5 • 0.100
pH = 2.87 • 2.) What is the pH after 10.0ml of NaOH have been added? • 25.0ml x 1 liter x 0.100 mol = 0.0025mol HC2H3O2 • 1000ml 1 liter • 10.0ml x 1 liter x 0.100 mol = 0.00100mol OH- • 1000ml 1 liter
HC2H3O2 + NaOH NaC2H3O2 + HOH • HC2H3O2 + OH-C2H3O2-+HOH • .0025 .00100 0 • -.00100 -.00100 +.00100 • 0.0015 0 +.00100 • USE THE HENDERSON-HASSELBALCH EQUATION
pH = pKa + log([CB]/[Acid]) • = -log(1.8 x 10-5) + log(.001/.0015) • = 4.74 + (-0.176) • = 4.56 • 3.) …after 12.5ml of 0.100M NaOH added? • HC2H3O2 + OH- C2H3O2-+HOH
12.5ml x 1 liter x 0.100 mol = 0.00125mol OH- 1000ml 1 liter • HC2H3O2 + OH- C2H3O2-+HOH • .0025 .00125 0 • -.00125 -.00125 +.00125 • .00125 0 +.00125 • pH = pKa + log(.00125/.00125) • = 4.74
pH = pKa @ ½ eq point • Law of COSINE • c2 = a2 + b2 – 2ab cos C • cos 90o = • Zero • c2 = a2 + b2 • Leads to the Pythagorean Theorem
4.) ….after 25.0ml added? • @ EQ point. • pH of a salt…. • 1.0x 10-14 = x2 • 1.8 x 10-5 .10/2 • pH = 8.72
5.) …after 26.0ml added? • (strong base in XS) • Do not use the HENDERSON-HASSELBALCH EQUATION • Use only when the XS is WEAK. • 25.0ml x 1 liter x 0.100 mol = 0.0025mol HC2H3O2 • 1000ml 1 liter • 26.0ml x 1 liter x 0.100 mol =0.00260 mol OH- • 1000ml 1 liter
HC2H3O2+ OH- C2H3O2-+H2O • 0.0025mol .0026 0 0 • -.0025 -.0025 +.0025 • 0 .0001 • pOH = -log (.0001mol/.051 liters) • = 2.71 • pH = 11.29
6.) …after 15.0ml added? • 25.0ml x 1 liter x 0.100 mol = 0.0025mol HC2H3O2 • 1000ml 1 liter • 15.0ml x 1 liter x 0.100 mol = 0.00150 mol OH- • 1000ml 1 liter • HC2H3O2+ OH- C2H3O2-+H2O • .0025mol .0015 0 0 • -.0015 -.0015 +.0015 • .0010 0 +.0015
pH = pKa + log([CB]/[Acid]) • = - log(1.8 x 10-5 ) + log(.0015mol/.001mol) • = 4.74 + .176 • = 4.92
17.4 Solubility Equilibrium • Ksp = solubility product • BaSO4 (s) Ba2+(aq) + SO42-(aq) • Ksp = [Ba2+][SO42-] called expression [BaSO4] Ca3(PO4)2 3Ca2+ + 2PO43- Ksp = [Ca2+ ]3[PO43-]2
Calculate the Ksp of Ag2SO4 if the silver conc is 0.00013 M. • Ag2SO4 2Ag+ + SO42- • ______ .00013M .000065M • Ksp = [Ag+]2[SO42-] • = (1.3 x 10-4)2(.65 x 10-4) • = 1.1 x 10-12
What is the solubility of CaF2 in g/l if the Ksp is 3.9 x 10-11 • CaF2 Ca2+ + 2F- • x x 2x • Ksp = [Ca2+ ][F- ]2 • 3.9 x 10-11 = (x)(2x)2 = 4x3 • x = 2.1 x 10-4 M x 78.1 g CaF2 = 1.6 x 10-2 g/l • 1 mol
Will a ppt form? • Sample 17.15 • 0.10 L of 0.0080 M Pb(NO3)2 is mixed w/ 0.40 L of 0.0050 M of Na2SO4. Will a ppt form? • Pb(NO3)2 +Na2SO4 PbSO4+ NaNO3 • 0.10 L x 0.0080 mol/L = 8.0 x 10-4 mol Pb2+ • [Pb2+] = 8.0 x 10-4 mol = 1.6 x 10-3M • 0.50 L
0.40 L x 0.0050 mol/L = 2.0 x 10-3 mol SO42- • [SO42-] = 2.0 x 10-3 mol = 4.0 x 10-3M • 0.50 L • Ksp = [Pb2+] [SO42-] = 6.3 x 10-7 • = (1.6 x 10-3M)(4.0 x 10-3M) • Q = 6.4 x 10-6 • ppt will form!
Q • If Q > Ksp …then ppt. will form. • If Q < Ksp …then no ppt. • If Q = Ksp…then @ equilibrium.
• 17.5 Common ion effect • The presence of either Ca2+ (aq) or F- (aq) in a solution reduces the solubility equilibrium of CaF2 to the left. • The solute conc. decreases by the presence of a 2nd solute w/ a common ion.
Sample 17.12 • Calculate the molar solubility (M) of CaF2 at 25oC in a solution that is • a.) 0.010M in Ca(NO3)2 • Ca(NO3)2 Ca2+ + 2NO3- • 0.010 0.010 0.020 • Ksp = 3.9 x 10-11 Ksp = [Ca2+][F-]2 • Ksp = 3.9 x 10-11 = (0.010)(2x)2 • x = 3.1 x 10-5M
b.) 0.010M in NaF • NaF Na+ + F- • Ksp = 3.9 x 10-11 Ksp = [Ca2+][F-]2 • Ksp = 3.9 x 10-11 = (x)(0.01)2 • x = 3.9 x 10-7M