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Chapter 9: Q25P (page 209)

Show that the limiting fraction of association of a base in water, as the concentration of base approaches $0, is a = 10^{7} Kb / (1 + 10^{7} K_{b})$ Find the limiting value of $a$ for $K_{b} = 10^{- 4}$ and for $K_{b} = 10^{- 10}$

Short Answer

Expert verified

The $limiting value of 伪 for K_{b} = 10^{- 4} and for K_{b} = 10^{- 10}$ is 0.999 and 0.000999 respectively.

Step by step solution

01

Showing statement is true.

When bases approaches , $pH = 7 and [{OH}^{-}] = 10^{- 7} M$

$K_{b} = \frac{[{BH}^{+}] [{OH}^{-}]}{[B]} = \frac{[{BH}^{+}]}{[B] 脳 10^{- 7}} [{BH}^{+}] = 10^{7} K_{b} [B]$

Showing that the statement is true:

$伪 = \frac{[{BH}^{+}]}{[B] + [{BH}^{+}]} = \frac{10^{7} K_{b} [B]}{[B] + 10^{7} K_{b} [B]} = 10^{7} K_{b} / (1 + 10^{7} K_{b})$

02

Finding limiting value α for Kb=10-4and for Kb=10-10:

$a) for K_{b} = 10^{- 4} 伪 = 10^{7} K_{b} / (1 + 10^{7} K_{b}) = 10^{7} 脳 10^{- 4} / (1 + 10^{7} 脳 10^{- 4}) = 0.999 b) for K_{b} = 10^{- 10} 伪 = 10^{7} K_{b} / (1 + 10^{7} K_{b}) = 10^{7} 脳 10^{(} - 10) / (1 + 10^{7} 脳 10^{- 10}) = 0.000999$

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Most popular questions from this chapter

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(a) Calculate the pH of a solution prepared by mixing $0.00100 mol$ of the base $B (K_{b} = 10^{- 2.00})$ with $0.0200 mol$ of ${BH}^{+} {Br}^{-}$ and diluting to $1.00 L$ .First calculate the pH by assuming $[B] = 0.0100$ and $[{BH}^{+}] = 0.0200 M$ . Compare this answer with the pH calculated without making such an assumption.

(b) After working part (a) by hand, use Excel Goal Seck to find the same answers.

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