Explanation The equilibrium reaction is given below. C 5 H 5 N ( a q ) + H 2 O ( l ) ⇌ C 5 H 5 N H + ( a q ) + O H − ( a q ) The base ionization constant can be written as given below. K b = [ C 5 H 5 N H + ] [ O H − ] [ C 5 H 5 N ] The value of K b can be calculated as given below. K b = K w K a = 1.00 × 10 − 14 6.3 × 10 − 6 = 1.6 × 10 − 9 . The p H of the solution is 8.5. So the concentration of hydroxide ion can be calculated as shown below. [ O H − ] = 10 − p O H = 10 − ( 14.00 − p H ) = 10 − ( 14.00 − 8.5 ) = 10 − 5.5 = 3.0 × 10 − 6 M . A table can be set up as shown below. C 5 H 5 N ( a q ) O H − ( a q ) C 5 H 5 N H + ( a q ) I n i t i a l c o n c . ( M ) 0.2 u n k n o w n 0 C h a n g e i n c o n c

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ISBN: 9781285199047

Author: John W. Moore, Conrad L. Stanitski

Publisher: Cengage Learning

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Chapter 14, Problem 63QRT

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Interpretation:

The concentration of unreacted pyridine in the solution has to be calculated.

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Chapter 14, Problem 62QRT

Chapter 14, Problem 64QRT

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LTS Solid: AT=Te-Ti Trial 1 Trial 2 Trial 3 Average ΔΗ Mass water, g 24.096 23.976 23.975 Moles of solid, mol 0.01763 001767 0101781 Temp. change, °C 2.9°C 11700 2.0°C Heat of reaction, J -292.37J -170.473 -193.26J AH, kJ/mole 16.58K 9.647 kJ 10.85 kr 16.58K59.64701 KJ mol 12.35k Minimum AS, J/mol K 41.582 mol-k Remember: q = mCsAT (m = mass of water, Cs=4.184J/g°C) & qsin =-qrxn & Show your calculations for: AH in J and then in kJ/mole for Trial 1: qa (24.0969)(4.1845/g) (-2.9°C)=-292.37J qsin = qrxn = 292.35 292.37J AH in J = 292.375 0.2923kJ 0.01763m01 =1.65×107 AH in kJ/mol = = 16.58K 0.01763mol mol qrx Minimum AS in J/mol K (Hint: use the average initial temperature of the three trials, con Kelvin.) AS=AHIT (1.65×10(9.64×103) + (1.0 Jimai

Chapter 14 Solutions

Chemistry: The Molecular Science

Ch. 14.1 - Identify each molecule or ion as a Brnsted-Lowry...Ch. 14.1 - Using Le Chatelier’s Principle Use Le Chatelier’s...Ch. 14.1 - Prob. 14.3E Ch. 14.1 - Complete the table. Ch. 14.1 - Prob. 14.4E Ch. 14.1 - Prob. 14.5E Ch. 14.2 - Prob. 14.6CE Ch. 14.2 - Prob. 14.7E Ch. 14.3 - Prob. 14.2PSP Ch. 14.3 - Prob. 14.3PSP

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The concentration of unreacted pyridine in the solution has to be calculated.

Solution Summary: The author explains that the concentration of unreacted pyridine in the solution is 0.2.

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Interpretation: The concentration of unreacted pyridine in the solution has to be calculated.

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Chapter 14 Solutions

Interpretation:

The concentration of unreacted pyridine in the solution has to be calculated.