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Unformatted text preview: estrada (gse77) – Homework 8 – Sutcliffe – (52410) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which equation represents K a2 for sulfurous acid? 1. HSO 3 (aq) + H 2 O( ℓ ) → H 2 SO 3 (aq) + OH (aq) 2. SO 2 3 (aq) + H 2 O( ℓ ) → HSO 3 (aq) + OH (aq) 3. H 2 SO 3 (aq) + H 2 O( ℓ ) → HSO 3 (aq) + H 3 O + (aq) 4. H 2 SO 3 (aq) + 2 H 2 O( ℓ ) → SO 2 3 (aq) + 2 H 3 O + (aq) 5. HSO 3 (aq) + H 2 O( ℓ ) → SO 2 3 (aq) + H 3 O + (aq) correct Explanation: 002 10.0 points The pH of 0.010 M H 3 PO 4 (aq) is 2.24. Esti mate the concentration of HPO 2 4 in the solu tion. For H 3 PO 4 , the values of K a1 , K a2 , and K a3 are 7 . 6 × 10 3 , 6 . 2 × 10 8 , and 2 . 1 × 10 13 , respectively. 1. 5 . 8 × 10 3 M 2. 6 . 2 × 10 8 M correct 3. 7 . 6 × 10 3 M 4. 0.010 M 5. 2 . 1 × 10 13 M Explanation: 003 10.0 points Oxalic acid is a relatively strong diprotic acid with K 1 = 5 . 3 × 10 2 and K 2 = 5 . 3 × 10 5 . What is the [HC 2 O 4 ] concentration in an ox alic acid solution that is 0.10 M? 1. 0.073 M 2. 0.051 M correct 3. 0.027 M 4. 0.10 5. 5 . 3 × 10 5 Explanation: K 1 = 5.3 × 10 2 K 2 = 5.3 × 10 5 In this case, since we’re considering a solu tion of H 2 C 2 O 4 and HC 2 O 4 , we should use the expression for K 1 here: K 1 = [H + ] [HC 2 O 4 ] [H 2 C 2 O 4 ] Substituting in values for the concentrations and K 1 , we have 5.3 × 10 2 = ( x ) ( x ) (0 . 10 x ) As H 2 C 2 O 4 has a relatively large K 1 , we CANNOT assume that x is very small com pared to 0.10 and consequently we must use the quadratic equation to solve for x : 5 . 3 × 10 2 (0 . 10 x ) = x 2 . 0053 5 . 3 × 10 2 x = x 2 x 2 + 5 . 3 × 10 2 x . 0053 = 0 Using the quadratic equation, we discover that x is 0.051. Bearing in mind that we called [HC 2 O 4 ] x in the equation, [HC 2 O 4 ] is 0.051 M. 004 10.0 points What is the pH of a 0 . 21 M solution of benzy lammonium chloride (C 6 H 5 CH 2 NH 3 Cl)? K b for benzylamine is 2 . 2 × 10 5 . Your answer must be within ± 0.4% Correct answer: 5 . 0101. Explanation: M C 6 H 5 CH 2 NH 3 Cl = 0 . 21 M K b = 2 . 2 × 10 5 It’s a salt of a weak base (BHX). This means you need a K a for the weak acid BH + : K a = K w K b = 1 . × 10 14 2 . 2 × 10 5 estrada (gse77) – Homework 8 – Sutcliffe – (52410) 2 = 4 . 54545 × 10 10 You CAN use the approximation for the equilibrium which means that [H + ] = radicalbig K a · C BH + = radicalBig (4 . 54545 × 10 10 ) (0 . 21) = 9 . 77008 × 10 6 M pH = log(9 . 77008 × 10 6 ) = 5 . 0101 005 10.0 points Note this is a SALT.. think about what it is made up of and what that will do in water!...
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This note was uploaded on 03/03/2012 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Holcombe

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