Chapter 10: Q28P (page 232)

A dibasic compound, B, has pKb1 5 4.00 and pKb2 5 6.00.
Find the fraction in the form ${BH}_{2}^{2 +}$ at pH 7.00, using Equation
10-19. Note that K1 and K2 in Equation 10-19 are acid dissociation
constants forlocalid="1655450143786" $^{{BH}_{2}^{2 +}}$
1 (K1 = Kw/Kb2 and K2 = Kw/Kb1).

Short Answer

Expert verified

The answer is $a ({BH}_{2}^{2 +}) = 0.91$ .

Step by step solution

Dibasic acid

Dibasic, or diprotic acid, an acid containing two potential protons to donate. Dibasic salt, is a salt with two hydrogen atoms, with respect to the parent acid, replaced by cations. Dibasic ester, is an ester of a dicarboxylic acid.

Calculating fraction in the form at pH 7.00

We will use the following:

$^{[H^{+}] = 10^{- 7}}$ from $^{pH = 7.00}$

$^{K_{1} = 10^{- 8}}$ from $^{{pK}_{b 2} = 6.00}$

$^{K_{2} = 10^{- 10}}$ from $^{p_{b 1} = 4.00}$

In order to find the fraction in the form of $^{{BH}_{2}^{2 +}}$ ,

$a ({BH}_{2}^{2}) = \frac{[H^{+}]}{{[H^{+}]}^{2} + [H^{+}] K_{1 +} K_{1} K_{2}} a ({BH}_{2}^{2}) = \frac{{(10^{- 7})}^{2}}{{(10^{- 7})}^{2} + (10^{- 7}) \times (10^{- 8}) + (10^{- 8} \times 10^{- 10})} a ({BH}_{2}^{2}) = 0.91$

The final answer is; $^{a ({BH}_{2}^{2}) = 0.91}$

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A dibasic compound, B, has pKb1 5 4.00 and pKb2 5 6.00.
Find the fraction in the form ${BH}_{2}^{2 +}$ at pH 7.00, using Equation
10-19. Note that K1 and K2 in Equation 10-19 are acid dissociation
constants forlocalid="1655450143786" $^{{BH}_{2}^{2 +}}$
1 (K1 = Kw/Kb2 and K2 = Kw/Kb1).

Short Answer

Step by step solution

Dibasic acid

Calculating fraction in the form at pH 7.00

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A dibasic compound, B, has pKb1 5 4.00 and pKb2 5 6.00.Find the fraction in the form BH22+at pH 7.00, using Equation10-19. Note that K1 and K2 in Equation 10-19 are acid dissociationconstants forlocalid="1655450143786" BH22+1 (K1 = Kw/Kb2 and K2 = Kw/Kb1).

Short Answer

Step by step solution

Dibasic acid

Calculating fraction in the form at pH 7.00

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Chemistry Branches

Ionic and Molecular Compounds

Kinetics

Organic Chemistry

Chemical Reactions

Study anywhere. Anytime. Across all devices.

A dibasic compound, B, has pKb1 5 4.00 and pKb2 5 6.00.
Find the fraction in the form ${BH}_{2}^{2 +}$ at pH 7.00, using Equation
10-19. Note that K1 and K2 in Equation 10-19 are acid dissociation
constants forlocalid="1655450143786" $^{{BH}_{2}^{2 +}}$
1 (K1 = Kw/Kb2 and K2 = Kw/Kb1).