Explanation Explanation The [ OH − ] of given NH 3 is 4 . 1 5 × 1 0 - 4 M _ . The compound NH 3 is a weak base and establishes equilibrium upon dissociation in an aqueous solution. The [ OH − ] value is calculated by the formation of an ICE (Initial Concentration Equilibrium) table. NH 3 + H 2 O ⇌ NH 4 + + OH − Initial ( M ) : 0.10 0 0 Change ( M ) : − x x x Equilibrium ( M ) : 0.10 − x x x The equilibrium constant expression is calculated by the formula, K b = [ Concentration of products ] [ Concentration of reactants ] The equilibrium constant expression for the stated reaction is, K b = [ NH 4 + ] [ OH − ] [ NH 3 ] The value of K b is 1.8 × 10 − 5 . Substitute the value of K b , [ NH 4 + ] , [ OH − ] and [ NH 3 ] from ICE table in the above expression. 1.8 × 10 − 5 = x × x 0.10 − x x 2 + 1.8 × 10 − 5 − 1.8 × 10 − 7 = 0 x = 4 . 1 5 × 1 0 - 4 M _ Thus, the value of [ OH − ] is 4 . 1 5 × 1 0 - 4 M _ . The pOH of given NH 3 solution is 3 . 3 8 _ . The pOH value is calculated by the formula, pOH = − log [ OH − ] Substitute the value of [ OH − ] in the above formula. pOH = − log [ 4.15 × 10 − 4 ] pOH = 3 . 3 8 _ The pH of given NH 3 is 1 0 . 6 2 _ . The sum, pH + pOH = 14 Substitute the calculated value of pOH in the above equation. pH + 3.38 = 14 pH = 14 − 3.38 pH = 1 0 . 6 2 _ The pH of KCl is 7 _ . The strong salts like KCl are formed by the reaction between a strong acid and a strong base. KCl is made up by the combination of a strong acid, that is HCl and a strong base, that is KOH . These are strong salts and dissociate completely in an aqueous. The strong salts like KCl are neutral in nature and have pH value equal to 7 _ . The [ OH − ] value of KCl is. 0 . 1 0 M _ . The given compound KOH is a strong base and dissociates completely and readily in an aqueous solution. The dissociation reaction that takes place is, KOH → K + + OH − Thus, the given concentration of KOH will correspond to the [ OH − ] . Therefore, the value of [ OH − ] is 0 . 1 0 M _ . The pOH of given KCl solution is 1 _ . The pOH value is calculated by the formula, pOH = − log [ OH − ] Substitute the value of [ OH − ] in the above formula

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Chapter 13, Problem 166CWP

Interpretation Introduction

Interpretation: The given solutions are to be arranged in increasing order of pH.

Concept introduction: Strong acids dissociate completely and rapidly when dissolved in an aqueous solution and have a very large value of Ka and a lower value of pH while the weak acids dissociate partially and slowly in an aqueous solution and have a very low value of Ka.and a higher value of pH.

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Chapter 13, Problem 165CWP

Chapter 13, Problem 167CWP

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Chapter 13 Solutions

EBK CHEMISTRY: AN ATOMS FIRST APPROACH

Ch. 13 - Define each of the following: a. Arrhenius acid b....Ch. 13 - Define or illustrate the meaning of the following...Ch. 13 - Prob. 3RQ Ch. 13 - How is acid strength related to the value of Ka?...Ch. 13 - Two strategies are followed when solving for the...Ch. 13 - Prob. 6RQ Ch. 13 - Prob. 7RQ Ch. 13 - For conjugate acidbase pairs, how are Ka and Kb...Ch. 13 - What is a salt? List some anions that behave as...Ch. 13 - For oxyacids, how does acid strength depend on a....

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Solution Summary: The author explains that the given solutions are to be arranged in increasing order of pH. Strong acids dissociate completely and rapidly when dissolved in an aqueous solution.

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