Chapter 6: Q49P (page 143)

Write the $K_{b}$ reaction of ${CN}^{-}$ Given that the $K_{a}$ value for $HCN$ is $6.2 \times 10^{- 10}$ calculate $K_{b}$ for ${CN}^{-}$ .

Short Answer

Expert verified

The reaction is given below.

${CN}^{-} + H_{2} O \overset{K_{b}}{⇇} HCN + {OH}^{-} {CN}^{-} + H_{2} O \overset{K_{b}}{⇇} HCN + {OH}^{-}$

The value of $K_{b} is 1.6 \times 10^{- 5}$ .

Step by step solution

Acid dissociation constant

A weak acid maintains equilibrium with its ions in aqueous solution. Since equilibrium is established, there must be an equilibrium constant for the reaction. The acid dissociation constant is given by the following relation:

$HA + H_{2} O ⇌ A^{-} + H_{3} O^{+}$

$K_{a} = \frac{[A^{-}] [H^{+}]}{[HA]}$

Base dissociation constant

A weak base remains in equilibrium with its ions in an aqueous solution.

$BOH + H_{2} O ⇌ {BH}^{+} + {OH}^{-}$

The equilibrium constant is known as the base dissociation constant, and it is given by the following relations.

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

Step 3: KB reaction for CN-

is the conjugate base of HCN. The acid-base reaction is given below.

${CN}^{-} + H_{2} O \overset{K_{b}}{⇇} HCN + {OH}^{-}$

Calculation of Kb

The relation between $K_{a}$ and $K_{b}$ is:

$K_{w} = K_{a} \times K_{b} K_{b} = \frac{K_{w}}{K_{a}} = \frac{1.0 \times 10^{- 14}}{6.2 \times 10^{- 10}} = 1.6 \times 10^{- 5}$

The value of $K_{b} is 1.6 \times 10^{- 5}$ .

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Write the $K_{b}$ reaction of ${CN}^{-}$ Given that the $K_{a}$ value for $HCN$ is $6.2 \times 10^{- 10}$ calculate $K_{b}$ for ${CN}^{-}$ .

Short Answer

Step by step solution

Acid dissociation constant

Base dissociation constant

Step 3: KB reaction for CN-

Calculation of Kb

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Write the Kbreaction of CN- Given that the Ka value for HCNis6.2×10-10 calculateKb forCN-.

Short Answer

Step by step solution

Acid dissociation constant

Base dissociation constant

Step 3: KB reaction for CN-

Calculation of Kb

One App. One Place for Learning.

Most popular questions from this chapter

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Write the $K_{b}$ reaction of ${CN}^{-}$ Given that the $K_{a}$ value for $HCN$ is $6.2 \times 10^{- 10}$ calculate $K_{b}$ for ${CN}^{-}$ .