CH302 Homework 5-solutions

# CH302 Homework 5-solutions - Bodet(ngb299 Homework 5...

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Bodet (ngb299) – Homework 5 – sutclife – (51630) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. Please note: ’p’ just equals ’negative log o±’ so we can have pH, pOH , pK a etc. This HW combined with HW 3 4 will cover us up to exam 2, so salts, neutralizations, bufers and indicators will be on Exam 3 (that’s section 7.8, etc.) HW due the evening be±ore the exam. 001 10.0 points How many grams o± HNO 3 are needed to give apHo ±3 . 75 in an 11 . 5Ltank? Themo lar mass o± HNO 3 is 63 . 1g/mol. 1. 0.0112 g 2. 0.000178 g 3. 0.0000324 g 4. 1.36 g 5. 0.00204 g 6. 0.466 g 7. 5.49 g 8. 0.129 g correct 9. 2.19 g Explanation: pH = 3 . 75 V =11 . 5L MM = 63 . 1g/mol [H + ]=10 - pH =10 - 3 . 75 =0 . 000177828 Since HNO 3 is a strong acid, HNO 3 (aq) H + (aq) + NO - 3 (aq) . One HNO 3 one H + ,so [H + ]=[HNO 3 ]=0 . 000177828 M and (11 . 5L) × 0 . 000177828 mol HNO 3 1L . 00204502 mol HNO 3 . Thus (0 . 00204502 mol HNO 3 ) × 63 . 1gHNO 3 1mo lHNO 3 . 129041 g HNO 3 002 10.0 points The pH o± 0 . 1MHC lO 2 (chlorous acid) aqueous solution was measured to be 1 . 2. What is the value o± p K a ±or chlorous acid? 1. 3.91 2. 2.57 3. 0.11 4. 1.20 5. 0.96 correct 6. 1.40 Explanation: M . 1M pH=1 . 2 Analyzing the reaction with molarities, HClO 2 +H 2 O ± ² H 3 O + +ClO - 2 0 . 1 - 00 - x - xx 0 . 1 - x - [H 3 O + ]=[ClO - 2 - pH - 1 . 2 . 0630957 mol / L The K a is K a = [H 3 O + ][ClO - 2 ] [HClO 2 ] = (0 . 0630957) 2 0 . 1 - 0 . 0630957 . 107876 and the p K a is p K a = - log(0 . 107876) = 0 . 967077 .

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Bodet (ngb299) – Homework 5 – sutclife – (51630) 2 003 10.0 points A7 . 2 × 10 - 3 MsolutionoFaceticacidis5.0% dissociated. In a 7 . 2 × 10 - 4 Mso lut ion ,the percent dissociation would be 1. zero. 2. < 5%. 3. None oF these 4. the same. 5. > 5%. correct Explanation: 004 (part 1 of 3) 10.0 points AgivenweakacidHZhasa K a =3 . 1 × 10 - 6 . What is the H 3 O + concentration oF a solution oF HZ that has a concentration oF 0 . 07 mol/L? Correct answer: 0 . 000465833 mol / L. Explanation: K a . 1 × 10 - 6 [HZ] = 0 . 07mol / L The expression For K a here is K a = [H + ][Z - ] [HZ] Substituting in values For the concentrations and K a ,wehave 3 . 1 × 10 - 6 = ( x )( x ) (0 . 07 - x ) As HZ has a small K a ,w ecana s sumetha t x is very small compared to 0 . 07, so we can
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CH302 Homework 5-solutions - Bodet(ngb299 Homework 5...

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