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Chapter 6: Q6TY (page 138)

$K_{b}$ for dimethylamine is $5.9 脳 10^{- 4}$ . Find $K_{a}$ for dimethyl ammonium ion.

Short Answer

Expert verified

For the dimethylammonium ion, the value of $K_{a} is 1.7 脳 10^{- 11}$ .

Step by step solution

01

Step 1: Define the formula for calculation of Kb

The value of $K_{b}$ can be calculated as

$K_{b} = \frac{[{BH}^{+}] [OH]}{[B]}$

where $K_{b}$ is called the base dissociation constant.

The value of $K_{w}$ is calculated by the formula

$K_{w} = K_{a} 脳 K_{b}$

02

Calculate the value of Kb

It is given that the value of $K_{b}$ for dimethylamine is $5.9 脳 10^{- 4}$ .

$K_{a}$ for the dimethylammonium ion can be derived from

$K_{w} = K_{a} 脳 K_{b}$

where

$(K_{b})$ is base dissociation constant

$(K_{w})$ is ionic product of water , and

$(K_{a})$ is acid dissociation constant.

$K_{a} = \frac{K_{w}}{K_{b}}$

$\begin{array}{l} = \frac{10 脳 10^{a}}{5.9 脳 10} \\ = 0.169 脳 10^{- 10} \\ = 1.7 脳 10^{- 11} \end{array}$

Thus, for the dimethylammonium ion, $K_{a} is 1.7 脳 10^{- 11}$ .

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Most popular questions from this chapter

The equilibrium constant for the reaction $H_{2} O 鈬� H^{+} + O H^{-}$ is $1.0 脳 10^{- 14}$ at $25 掳 C$ . What is the value of K for the reaction $4 H_{2} O 鈬� 4 H^{+} + 4 O H^{-}$ ?

Use Table 6-1 to find the $[H^{+}]$ in water at $100 掳 C$ and at $0 掳 C$ .

Find $[{Cu}^{2 +}]$ in a solution saturated with ${Cu}_{4} {(OH)}_{6} ({so}_{4})$ if $[{OH}^{-}]$ is fixed at $1.0 脳 10^{- 6} M$ . Note that ${Cu}_{4} {(OH)}_{6} ({so}_{4})$ gives 1mol of ${SO}_{4}^{2 -}$ 4 mol for of ${Cu}^{2 +}$ .

${Cu}_{4} {(OH)}_{6} ({so}_{4}) (s) 鈬� 4 C u^{2 +} + 6 O H^{-} + S O_{4}^{2 -} K_{s p} = 2.3 脳 10^{- 69}$

Question: (a) From K_w in Table 6-1, calculate the pH of pure water at 0⁰,20⁰, and 40⁰C.

(b) For the reaction , at 25⁰C. In this equation, D stands for deuterium, which is the isotope ²H. What is the pD (=-log[D⁺]) for neutral D₂O?

Find $[{La}^{3 +}]$ in the solution when excess solid lanthanum iodate, $La {({IO}_{3})}_{3}$ , is stirred with until the system reaches equilibrium. Assume that ${IO}_{3}^{-}$ from ${La}_{3} {({IO}_{3})}_{3}$ is negligible compared with ${IO}_{3}^{-}$ from ${LiOO}_{3}$ .

See all solutions

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