Lecture 9, Chemistry of Solutions notes
Lecture 9, Chemistry of Solutions notes 202-NYB-05
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This 4 page Class Notes was uploaded by CatLover44 on Sunday October 9, 2016. The Class Notes belongs to 202-NYB-05 at Dawson Community College taught by Nadia Schoonhoven in Fall 2016. Since its upload, it has received 5 views. For similar materials see Chemistry of Solutions in Chemistry at Dawson Community College.
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Date Created: 10/09/16
Chemistry of Solutions Lecture 9 Thursday, September 22, 2016 Topic covered: calculating the pH of weak acidic solutions. The pH of Weak Acidic Solutions -4 Example: calculate the pH of a 1.00 M HF solution, whose K = 7.2xa0 . + - HF ⟷ H + F K a [H ] [F ] / [HF] 7.2x10 = x / 1.00 -4 1/2 -2 (7.2x10 ) = 2.7x10 = x + + x = [H ], pH = -log[H ] = 1.57. Checking the approximation: ( (2.7x10 ) / 1.00 ) x 100% = 2.7%, approximation is ok. Solving Weak Acid Equilibrium Problems + 1. List the major contributors of H . + 2. Choose the species that produce H , and write the balanced equations for their reactions with water. 3. Use the equilibrium constants for your reactions, and decide which reaction produces the + most H (which has the largest K ?).a + 4. Write the equilibrium expression for the dominant H donor. + 5. List the initial concentrations of the species in the reaction for the dominant H donor. 6. Define the value of x (change needed to reach equilibrium). 7. Write the equilibrium concentrations in terms of x and substitute the equilibrium concentrations into the equilibrium expression. 8. Approximate the value of x if K is mach smaller (at least 1000 times less) than the initial concentrations. If you can't approximate x, use the quadratic equation. 9. Calculate the equilibrium concentrations. 10.Use the 5% rule to verify whether the approximation was correct, and check your answers by calculating K using the equilibrium concentrations you calculated. + 11.Calculate [H ] and pH. + 12.Calculate the concentration of the other species in the solution using the [H ] . eq -8 Exercise: calculate the pH of a 0.100 M HOCl with K = 3.5x10a. Major components: HOCl, and H O. 2 HOCl H+ OCl - Initial 0.100 0 0 Change 0.1 – x 0 + x 0 + x -2 -5 -5 Equilibrium 9.9x10 5.9x10 5.9x10 + - HOCl ⟷ H + OCl H 2 ⟷ H + OH - -8 + - K a 3.5x10 = ( [H ] [OCl ] / [HOCl] ) -8 2 3.5x10 = x / 0.100 ((3.5x10 )(0.100)) 1/2= 5.9x10 = x -5 x = [H+] = 5.9x10 M pH = -log(5.9x10-5) = 4.23. -10 Exercise: calculate the pH of a solution that contains 1.00 M HCN, K = 6.2x10 a and 5.00 M HNO , 2 = a.0x10 , solution. What is the concentration of CN at equilibrium? 2 The major components are HCN, HNO , and H O. 2 2 + - -10 HCN ⟷ H + CN , K = 6.2x1a + - -4 HNO ⟷ 2 + NO , K = 4.0x10 a H 2 ⟷ H + OH , K = 1.0x1a -14 -4 + - K a 4.0x10 = ( [H ] [CN ] ) / [HCN] -4 2 4.0x10 = x / 5.00 (4.0x10 )-4 1/2= x x = 4.5x10 M =2 pH = -log(4.5x10-2) = 1.35 Percent Dissociation Percent dissociation = ( amount dissociated (M) / initial concentration (M) ) x 100% Generally, the percent dissociation increases as a weak acid becomes more dilute. + A greater [H ] implies a lower pH, which means the % dissociation is lower. A higher % dissociation implies a lower [H ], which implies a greater pH. 3 Exercise: compare the percent dissociation for a 2.00 M and a 0.300 M acetic acid. Ka = 1.8x10 . -5 CH 3OOH H + CH 3OO - Initial 2.00 0 0 Change 2 – x 0 + x 0 + x Equilibrium 1.994 0.006 0.006 + - CH 3OOH H CH 3OO Initial 0.300 0 0 Change 0.3 – x 0 + x 0 + x Equilibrium 0.298 M 0.0023 M 0.0023 M Solution #1: 1.8x10 = x / 2.00 x = [H ] = 0.006 M % dissociation = (0.006) / (2.00) x 100% = 0.30% pH = -log(0.006) = 2.22 -5 2 Solution #2: 1.8x10 = x / 0.300 + x = [H ] = 0.0023 M % dissociation = (0.0023) / (0.300) x 100% = 0.77% pH = -log(0.0023) = 2.64 Percent dissociation increased as [H ] decreased. 4
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