H06 Buffer Solution

# H06 Buffer Solution - patino(mp25752 – H06 Buffers –...

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Unformatted text preview: patino (mp25752) – H06: Buffers – McCord – (53215) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. You might want to wait on some of these questions until you have heard me lecture on the subject. You’ve been warned.- Dr. McCord 001 10.0 points Calculate the concentration of HCO − 3 present in 0 . 3274 M H 2 CO 3 . Correct answer: 0 . 000375209 mol / L. Explanation: K w = 1 × 10 − 14 The equilibrium reactions of interest are H 2 CO 3 (aq) + H 2 O( ℓ ) ⇀ ↽ H 3 O + (aq) + HCO − 3 (aq) K a1 = 4 . 3 × 10 − 7 HCO − 3 (aq) + H 2 O( ℓ ) ⇀ ↽ H 3 O + (aq) + CO 2 − 3 (aq) K a2 = 5 . 6 × 10 − 11 Because the second ionization constant is much smaller than the first, we can assume that the first step dominates. H 2 CO 3 + H 2 O ⇀ ↽ H 3 O + + HCO − 3 . 3274 − − x − + x + x . 3274 − x − x x Assuming that x ≪ . 3274 M, K a1 = [H 3 O + ][HCO − 3 ] [H 2 CO 3 ] 4 . 3 × 10 − 7 = x 2 . 3274 − x ≈ x 2 . 3274 x = radicalBig (0 . 3274) (4 . 3 × 10 − 7 ) = 0 . 000375209 M . Because x < 1% of 0 . 3274 M, the assump- tion was valid, so x = [H 3 O + ] = [HCO − 3 ] = 0 . 000375209 M . 002 10.0 points Scientists investigating acid rain measured the pH of a water sample from a lake and found it to be 4 . 8. The total concentra- tion of dissolved carbonates in the lake is . 0045 M. Determine the molar concentra- tion of H 2 CO 3 in the lake if K a1 = 4 . 3 × 10 − 7 and K a2 = 5 . 6 × 10 − 11 . Correct answer: 0 . 00438113 mol / L. Explanation: pH = 4 . 8 K a1 = 4 . 3 × 10 − 7 K a2 = 5 . 6 × 10 − 11 C ◦ carbonates = 0 . 0045 M [H 3 O + ] = 10 − pH = 10 − 4 . 8 = 1 . 58489 × 10 − 5 The total concentration is . 0045 = [CO 2 − 3 ] + [HCO − 3 ] + [H 2 CO 3 ] (1) The first equilibrium reaction is H 2 CO 3 (aq) + H 2 O( ℓ ) ⇀ ↽ HCO − 3 (aq) + H 3 O + K a1 = 4 . 3 × 10 − 7 K a1 = [H 3 O + ][HCO − 3 ] [H 2 CO 3 ] 4 . 3 × 10 − 7 = (1 . 58489 × 10 − 5 ) [HCO − 3 ] [H 2 CO 3 ] [HCO − 3 ] = (4 . 3 × 10 − 7 ) [H 2 CO 3 ] 1 . 58489 × 10 − 5 = 0 . 0271312[H 2 CO 3 ] (2) The second equilibrium reaction is HCO − 3 (aq) + H 2 O( ℓ ) ⇀ ↽ CO 2 − 3 (aq) + H 3 O + K a2 = 5 . 6 × 10 − 11 K a2 = [H 3 O + ][CO 2 − 3 ] [HCO − 3 ] 5 . 6 × 10 − 11 = (1 . 58489 × 10 − 5 ) [CO 2 − 3 ] [HCO − 3 ] [HCO − 3 ] = (1 . 58489 × 10 − 5 ) [CO 2 − 3 ] 5 . 6 × 10 − 11 = 2 . 83017 × 10 5 [CO 2 − 3 ] (3) Equating (2) and (3), . 0271312[H 2 CO 3 ] = 2 . 83017 × 10 5 [CO 2 − 3 ] patino (mp25752) – H06: Buffers – McCord – (53215) 2 [H 2 CO 3 ] = (1 . 04314 × 10 7 ) [CO 2 − 3 ] (4) Substituting equations (3) and (4) into equation (1), . 0045 = [CO 2 − 3 ] + 2 . 83017 × 10 5 [CO 2 − 3 ] + (1 . 04314 × 10 7 ) [CO 2 − 3 ] [CO 2 − 3 ] = . 0045 1 + 2 . 83017 × 10 5 + 1 . 04314 × 10 7 = 4 . 19994 × 10 − 10 mol / L ....
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H06 Buffer Solution - patino(mp25752 – H06 Buffers –...

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