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Chapter 6: Q5TY (page 138)

$K_{a}$ for chloroacetic acid is $1.36 \times 10^{- 3} .$ Find $K_{b}$ for chloroacetate ion.

Short Answer

Expert verified

In the chloroacetate ion the value of $K_{b} is 7.4 \times 10^{- 12}$

Step by step solution

01

Step 1: Define the formula for the 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:

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

02

Calculate the value of Kb.

Given:

$K_{a} = 1.36 \times 10^{- 3}$

$K_{b}$ for chloroacetate ion

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

Where,

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

$(K_{w})$ is the ionic product of water $(1.00 \times 10^{- 14})$

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

$K_{b} = \frac{K_{w}}{K_{a}} = \frac{1.0 \times 10^{- 14}}{1.36 \times 10^{- 3}} = = 0.735 \times 10^{- 11} = 7.4 \times 10^{- 12}$

Thus, in the chloroacetate ion, the value of role="math" localid="1663408479676" $K_{b} is 7.4 \times 10^{- 12}$

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Most popular questions from this chapter

Identify the Bronsted-Lowry acids among the reactants in the following reactions:

(a) $KCN + HI ⇌ HCN + KI$

(b) ${PO}_{4}^{3 -} + H_{2} O ⇌ {HPO}_{4}^{2 -} + {OH}^{-}$

Reaction 6-8 is allowed to come to equilibrium in a solution initially containing $0.0100 {MBrO}_{3}^{-}$ , $0.0100 {MCr}^{3 +}$ and 1.00MH⁺. To find the concentrations at equilibrium, we construct the table at the bottom of the page showing initial and final concentrations. We use the stoichiometry coefficients of the reaction to say that if $x m ol$ of ${Br}^{-}$ are created, then we must also make x mol of ${Cr}_{2} O_{7}^{2 -}$ and 8x mol of H⁺. To produce x mol of ${Br}^{-}$ , we must have consumed x mol of ${Br}^{-} O_{3}^{-}$ and 2x mol of Cr³⁺.

(a) Write the equilibrium constant expression that you would use to solve for x to find the concentrations at equilibrium. Do not try to solve the equation.

(b) Because $K = 1 \times 10^{11}$ , we suppose that the reaction will go nearly "to completion." That is, we expect both the concentration of ${Br}^{-}$ and ${Cr}_{2} O_{7}^{2 -}$ to be close to 0.00500M an equilibrium. (Why?) That is, $x \approx 0.00500 M$ . With this value of $x, [H^{+}] = 1.00 + 8 x = 1.04 M$ and $[{BrO}_{3}^{-}] = 0.0100 - x = 0.0050 M$ . However, we cannot say $[{Cr}^{3 +}] = 0.0100 - 2 x = 0$ , because there must be some small concentration of ${Cr}^{3 +}$ at equilibrium. Write $[{Cr}^{3 +}]$ for the concentration of ${Cr}^{3 +}$ and solve for $[{Cr}^{3 +}]$ . The limiting reagent in this example is ${Cr}^{3 +}$ . The reaction uses up before consuming ${BrO}_{3}^{-}$ .

Find $[{Pb}^{2 +}], {PbI}_{2} (aq),$ and $[{PbI}_{3}^{-}]$ in a saturated solution of ${PbI}_{2} (s)$ with $[I^{-}] = 0.10 M .$

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

Which will be more soluble (moles of metal dissolved per litre of solution), $B a {(I O_{3})}_{2} (K_{s p} = 1.5 \times 10^{- 9})$ or $C a {({IO}_{3})}_{2} (K_{sp} = 7.1 \times 10^{7})$ ? Give an example of a chemical reaction that might occur that would reverse the predicted solubilities.

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