solution8_pdf

# solution8_pdf - toupal (rgt374) Homework 8 Sutclie (52410)...

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toupal (rgt374) – Homework 8 – Sutclife – (52410) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. 001 10.0 points Which equation represents K a2 ±or sul±urous acid? 1. SO 2 - 3 (aq) + H 2 O( ) HSO - 3 (aq) + OH - (aq) 2. H 2 SO 3 (aq) + H 2 O( ) HSO - 3 (aq) + H 3 O + (aq) 3. HSO - 3 (aq) + H 2 O( ) SO 2 - 3 (aq) + H 3 O + (aq) correct 4. H 2 SO 3 (aq) + 2 H 2 O( ) SO 2 - 3 (aq) + 2 H 3 O + (aq) 5. HSO - 3 (aq) + H 2 O( ) H 2 SO 3 (aq) + OH - (aq) Explanation: 002 10.0 points The pH o± 0.010 M H 3 PO 4 (aq) is 2.24. Esti- mate the concentration o± HPO 2 - 4 in the solu- tion. ²or H 3 PO 4 , the values o± K a1 , K a2 , and K a3 are 7 . 6 × 10 - 3 , 6 . 2 × 10 - 8 , and 2 . 1 × 10 - 13 , respectively. 1. 7 . 6 × 10 - 3 M 2. 0.010 M 3. 5 . 8 × 10 - 3 M 4. 2 . 1 × 10 - 13 M 5. 6 . 2 × 10 - 8 M correct 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. 0.051 M correct 2. 5 . 3 × 10 - 5 3. 0.073 M 4. 0.10 5. 0.027 M Explanation: K 1 = 5.3 × 10 - 2 K 2 = 5.3 × 10 - 5 In this case, since we’re considering a solu- tion o± H 2 C 2 O 4 and HC 2 O - 4 , we should use the expression ±or K 1 here: K 1 = [H + ] [HC 2 O 4 - ] [H 2 C 2 O 4 ] Substituting in values ±or 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 ±or x : 5 . 3 × 10 - 2 (0 . 10 - x ) = x 2 0 . 0053 - 5 . 3 × 10 - 2 x = x 2 x 2 + 5 . 3 × 10 - 2 x - 0 . 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 o± a 0 . 33 M solution o± benzy- lammonium chloride (C 6 H 5 CH 2 NH 3 Cl)? K b ±or benzylamine is 2 . 2 × 10 - 5 . Your answer must be within ± 0.4% Correct answer: 4 . 91195. Explanation: M C 6 H 5 CH 2 NH 3 Cl = 0 . 33 M K b = 2 . 2 × 10 - 5 It’s a salt o± a weak base (BHX). This means you need a K a ±or the weak acid BH + : K a = K w K b = 1 . 0 × 10 - 14 2 . 2 × 10 - 5

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toupal (rgt374) – Homework 8 – Sutclife – (52410) 2 = 4 . 54545 × 10 - 10 You CAN use the approximation For the equilibrium which means that [H + ] = r K a · C BH + = R (4 . 54545 × 10 - 10 ) (0 . 33) = 1 . 22474 × 10 - 5 M pH = - log(1 . 22474 × 10 - 5 ) = 4 . 91195 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 05/01/2009 for the course CH 52410 taught by Professor Sutcliffe during the Spring '09 term at University of Texas.

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solution8_pdf - toupal (rgt374) Homework 8 Sutclie (52410)...

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