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Chapter 14: Q113 E (page 837)

Draw a curve for a series of solutions of HF. Plot \({\left[ {{H_3}{O^ + }} \right]_{total }}\) on the vertical axis and the total concentration of HF (the sum of the concentrations of both the ionized and nonionized HF molecules) on the horizontal axis. Let the total concentration of HF vary from \(1 \times 1{0^{ - 10}}M\) to\(1 \times 1{0^{ - 2}}M\) .

Short Answer

Expert verified

\(\begin{array}{*{20}{r}}{(HF) = 1 \times 1{0^{ - 10}}M/\left( {{H_3}{O^ + }} \right) = 1.001 \times 1{0^{ - 7}}M}\\{(HF) = 1 \times 1{0^{ - 9}}M/\left( {{H_3}{O^ + }} \right) = 1.01 \times 1{0^{ - 7}}M}\\{(HF) = 1 \times 1{0^{ - 8}}M/\left( {{H_3}{O^ + }} \right) = 1.1 \times 1{0^{ - 7}}M}\\{(HF) = 1 \times 1{0^{ - 7}}M/\left( {{H_3}{O^ + }} \right) = 2 \times 1{0^{ - 7}}M}\\{(HF) = 1 \times 1{0^{ - 6}}M/\left( {{H_3}{O^ + }} \right) = 1.1 \times 1{0^{ - 6}}M}\\{(HF) = 1 \times 1{0^{ - 5}}M/\left( {{H_3}{O^ + }} \right) = 1.01 \times 1{0^{ - 5}}M}\\{(HF) = 1 \times 1{0^{ - 4}}M/\left( {{H_3}{O^ + }} \right) = 8.13 \times 1{0^{ - 5}}M}\\{(HF) = 1 \times 1{0^{ - 3}}M/\left( {{H_3}{O^ + }} \right) = 4.42 \times 1{0^{ - 4}}M}\\{(HF) = 1 \times 1{0^{ - 2}}M/\left( {{H_3}{O^ + }} \right) = 17.04 \times 1{0^{ - 4}}M}\end{array}\)

Step by step solution

01

To find the concentration of HF varies in \({\bf{1 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}{\bf{M}}\)\(\)

The reaction

- \({K_a}\) of \(HF\) is \(3.5 \times 1{0^{ - 4}}\)

- Concentration of HF varies from\(1 \times 1{0^{ - 10}}M\)to\(1 \times 1{0^{ - 2}}M\)

Let us draw a curve for a series of solutions of \(HF\) (vertical axis \(\left[ {{H_3}{O^ + }} \right]\) and horizontal axis [HF] (ionized and nonionized) )

- Since the concentration of \({H_3}{O^ + }\) an ion in pure water is \(1 \times 1{0^{ - 7}},\) the initial concentration of \({H_3}{O^ + }\) in a given solution will be \(1 \times 1{0^{ - 7}}.\)

\([HF] = 1 \times 1{0^{ - 10}}M\)

\[HF(aq) \to \]

\({H_3}{O^ + }(aq)\)

\({F^ - }(aq)\)

Initial (M)

\(1.1{0^{ - 10}}\)

\(1.1{0^{ - 7}}\)

0

Change (M)

-X

+X

+X

Equilibrium(M)

\(1.1{0^{ - 10}} - X\)

\(1.1{0^{ - 7}} - X\)

X

\(\begin{array}{l}{K_a} = \frac{{\left[ {{H_3}{O^ + }} \right] \times \left[ {{F^ - }} \right]}}{{[HF]}}\\3.5 \times 1{0^{ - 4}} = \frac{{\left( {1 \times 1{0^{ - 7}} + x} \right) \times x}}{{1 \times 1{0^{ - 10}} - x}}\\3.5 \times 1{0^{ - 14}} - 3.5 \times 1{0^{ - 4}}x = 1 \times 1{0^{ - 7}}x + {x^2}\\0 = {x^2} + 3.5 \times 1{0^{ - 4}}x - 3.5 \times 1{0^{ - 14}}\\x = 1 \times 1{0^{ - 10}}\\\left[ {{H_3}{O^ + }} \right] = 1 \times 1{0^{ - 7}} + 1 \times 1{0^{ - 10}} = 1.001 \times 1{0^{ - 7}}\end{array}\)

02

To find the concentration of HF varies in \[{\bf{1}} \times {\left( {{\bf{10}}} \right)^{ - {\bf{9}}}}\] 

\([HF] = 1 \times 1{0^{ - 9}}M\)

\[HF(aq) \to \]

\({H_3}{O^ + }(aq)\)

\({F^ - }(aq)\)

Initial (M)

\(1.1{0^{ - 9}}\)

\(1.1{0^{ - 7}}\)

0

Change (M)

-X

+X

+X

Equilibrium (M)

\(1.1{0^{ - 9}} - X\)

\(1.1{0^{ - 7}} - X\)

X

\(\begin{array}{l}{K_a} = \frac{{\left[ {{H_3}{O^ + }} \right] \times \left[ {{F^ - }} \right]}}{{[HF]}}\\3.5 \times 1{0^{ - 4}} = \frac{{\left( {1 \times 1{0^{ - 7}} + x} \right) \times x}}{{1 \times 1{0^{ - 9}} - x}}\\3.5 \times 1{0^{ - 13}} - 3.5 \times 1{0^{ - 4}}x = 1 \times 1{0^{ - 7}}x + {x^2}\\0 = {x^2} + 3.5 \times 1{0^{ - 4}}x - 3.5 \times 1{0^{ - 13}}\\x = 1 \times 1{0^{ - 9}}\\\left[ {{H_3}{O^ + }} \right] = 1 \times 1{0^{ - 7}} + 1 \times 1{0^{ - 9}} = 1.01 \times 1{0^{ - 7}}\end{array}\)

03

To find the concentration of HF varies in \({\bf{1 \times 1}}{{\bf{0}}^{{\bf{ - 8}}}}{\bf{M}}\)

\([HF] = 1 \times 1{0^{ - 8}}M\)

\[HF(aq) \to \]

\({H_3}{O^ + }(aq)\)

\({F^ - }(aq)\)

Initial (M)

\(1.1{0^{ - 8}}\)

\(1.1{0^{ - 7}}\)

0

Change (M)

-X

+X

+X

Equilibrium (M)

\(1.1{0^{ - 8}} - X\)

\(1.1{0^{ - 7}} - X\)

X

\(\begin{array}{l}{K_a} = \frac{{\left[ {{H_3}{O^ + }} \right] \times \left[ {{F^ - }} \right]}}{{[HF]}}\\3.5 \times 1{0^{ - 4}} = \frac{{\left( {1 \times 1{0^{ - 7}} + x} \right) \times x}}{{1 \times 1{0^{ - 8}} - x}}\\3.5 \times 1{0^{ - 12}} - 3.5 \times 1{0^{ - 4}}x = 1 \times 1{0^{ - 7}}x + {x^2}\\0 = {x^2} + 3.5 \times 1{0^{ - 4}}x - 3.5 \times 1{0^{ - 12}}\\x = 1 \times 1{0^{ - 8}}\\\left[ {{H_3}{O^ + }} \right] = 1 \times 1{0^{ - 7}} + 1 \times 1{0^{ - 8}} = 1.1 \times 1{0^{ - 7}}\end{array}\)

04

To find the concentration of HF varies in \({\bf{1 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{M}}\)

\([HF] = 1 \times 1{0^{ - 7}}M\)

\[HF(aq) \to \]

\({H_3}{O^ + }(aq)\)

\({F^ - }(aq)\)

Initial (M)

\(1.1{0^{ - 7}}\)

\(1.1{0^{ - 7}}\)

0

Change (M)

-X

+X

+X

Equilibrium (M)

\(1.1{0^{ - 7}} - X\)

\(1.1{0^{ - 7}} - X\)

X

\(\begin{array}{l}{K_a} = \frac{{\left[ {{H_3}{O^ + }} \right] \times \left[ {{F^ - }} \right]}}{{[HF]}}\\3.5 \times 1{0^{ - 4}} = \frac{{\left( {1 \times 1{0^{ - 7}} + x} \right) \times x}}{{1 \times 1{0^{ - 7}} - x}}\\3.5 \times 1{0^{ - 11}} - 3.5 \times 1{0^{ - 4}}x = 1 \times 1{0^{ - 7}}x + {x^2}\\0 = {x^2} + 3.5 \times 1{0^{ - 4}}x - 3.5 \times 1{0^{ - 11}}\\x = 1 \times 1{0^{ - 7}}\\\left[ {{H_3}{O^ + }} \right] = 1 \times 1{0^{ - 7}} + 1 \times 1{0^{ - 7}} = 2 \times 1{0^{ - 7}}\end{array}\)

05

To find the concentration of HF varies in \({\bf{1 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{M}}\) 

\([HF] = 1 \times 1{0^{ - 6}}M\)

\[HF(aq) \to \]

\({H_3}{O^ + }(aq)\)

\({F^ - }(aq)\)

Initial (M)

\(1.1{0^{ - 6}}\)

\(1.1{0^{ - 7}}\)

0

Change (M)

-X

+X

+X

Equilibrium (M)

\(1.1{0^{ - 6}} - X\)

\(1.1{0^{ - 7}} - X\)

X

\(\begin{array}{l}{K_a} = \frac{{\left[ {{H_3}{O^ + }} \right] \times \left[ {{F^ - }} \right]}}{{[HF]}}\\3.5 \times 1{0^{ - 4}} = \frac{{\left( {1 \times 1{0^{ - 7}} + x} \right) \times x}}{{1 \times 1{0^{ - 6}} - x}}\\3.5 \times 1{0^{ - 10}} - 3.5 \times 1{0^{ - 4}}x = 1 \times 1{0^{ - 7}}x + {x^2}\\0 = {x^2} + 3.5 \times 1{0^{ - 4}}x - 3.5 \times 1{0^{ - 10}}\\x = 1 \times 1{0^{ - 6}}\\\left[ {{H_3}{O^ + }} \right] = 1 \times 1{0^{ - 7}} + 1 \times 1{0^{ - 6}} = 1.1 \times 1{0^{ - 6}}\end{array}\)

06

To find the concentration of HF varies in \({\bf{1 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}{\bf{M}}\)

\([HF] = 1 \times 1{0^{ - 5}}M\)

\[HF(aq) \to \]

\({H_3}{O^ + }(aq)\)

\({F^ - }(aq)\)

Initial (M)

\(1.1{0^{ - 5}}\)

\(1.1{0^{ - 7}}\)

0

Change (M)

-X

+X

+X

Equilibrium (M)

\(1.1{0^{ - 5}} - X\)

\(1.1{0^{ - 7}} - X\)

X

\(\begin{array}{l}{K_a} = \frac{{\left[ {{H_3}{O^ + }} \right] \times \left[ {{F^ - }} \right]}}{{[H\;F]}}\\3.5 \times 1{0^{ - 4}} = \frac{{\left( {1 \times 1{0^{ - 7}} + x} \right) \times x}}{{1 \times 1{0^{ - 5}} - x}}\\3.5 \times 1{0^{ - 9}} - 3.5 \times 1{0^{ - 4}}x = 1 \times 1{0^{ - 7}}x + {x^2}\\0 = {x^2} + 3.5 \times 1{0^{ - 4}}x - 3.5 \times 1{0^{ - 9}}\\x = 1 \times 1{0^{ - 5}}\\\left[ {{H_3}{O^ + }} \right] = 1 \times 1{0^{ - 7}} + 1 \times 1{0^{ - 5}} = 1.01 \times 1{0^{ - 5}}\end{array}\)

07

To find the concentration of HF varies in \({\bf{1 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{M}}\)

\([HF] = 1 \times 1{0^{ - 4}}M\)

\[HF(aq) \to \]

\({H_3}{O^ + }(aq)\)

\({F^ - }(aq)\)

Initial (M)

\(1.1{0^{ - 4}}\)

\(1.1{0^{ - 7}}\)

0

Change (M)

-X

+X

+X

Equilibrium (M)

\(1.1{0^{ - 4}} - X\)

\(1.1{0^{ - 7}} - X\)

X

\[\begin{array}{l}{K_a} = \frac{{\left[ {{H_3}{O^ + }} \right] \times \left[ {{F^ - }} \right]}}{{[HF]}}\\3.5 \times 1{0^{ - 4}} = \frac{{\left( {1 \times 1{0^{ - 7}} + x} \right) \times x}}{{1 \times 1{0^{ - 4}} - x}}\\3.5 \times 1{0^{ - 8}} - 3.5 \times 1{0^{ - 4}}x = 1 \times 1{0^{ - 7}}x + {x^2}\\0 = {x^2} + 3.5 \times 1{0^{ - 4}}x - 3.5 \times 1{0^{ - 8}}\\x = 8.12 \times 1{0^{ - 5}}\\\left[ {{H_3}{O^ + }} \right] = 1 \times 1{0^{ - 7}} + 8.12 \times 1{0^{ - 5}} = 8.13 \times 1{0^{ - 5}}\end{array}\]

08

To find the concentration of HF varies in \({\bf{1 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}{\bf{M}}\)

\([HF] = 1 \times 1{0^{ - 3}}M\)

\[HF(aq) \to \]

\({H_3}{O^ + }(aq)\)

\({F^ - }(aq)\)

Initial (M)

\(1.1{0^{ - 3}}\)

\(1.1{0^{ - 7}}\)

0

Change (M)

-X

+X

+X

Equilibrium (M)

\(1.1{0^{ - 3}} - X\)

\(1.1{0^{ - 7}} - X\)

X

\(\begin{array}{l}{K_a} = \frac{{\left[ {{H_3}{O^ + }} \right] \times \left[ {{F^ - }} \right]}}{{[HF]}}\\3.5 \times 1{0^{ - 4}} = \frac{{\left( {1 \times 1{0^{ - 7}} + x} \right) \times x}}{{1 \times 1{0^{ - 3}} - x}}\\3.5 \times 1{0^{ - 7}} - 3.5 \times 1{0^{ - 4}}x = 1 \times 1{0^{ - 7}}x + {x^2}\\0 = {x^2} + 3.5 \times 1{0^{ - 4}}x - 3.5 \times 1{0^{ - 7}}\\x = 4.42 \times 1{0^{ - 4}}\\\left[ {{H_3}{O^ + }} \right] = 1 \times 1{0^{ - 7}} + 4.42 \times 1{0^{ - 4}} = 4.42 \times 1{0^{ - 4}}\end{array}\)

09

To find the concentration of HF varies in \({\bf{1 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}{\bf{M}}\) 

\([HF] = 1 \times 1{0^{ - 2}}M\)

\[HF(aq) \to \]

\({H_3}{O^ + }(aq)\)

\({F^ - }(aq)\)

Initial (M)

\(1.1{0^{ - 2}}\)

\(1.1{0^{ - 7}}\)

0

Change (M)

-X

+X

+X

Equilibrium (M)

\(1.1{0^{ - 2}} - X\)

\(1.1{0^{ - 7}} - X\)

X

\(\begin{array}{l}{K_a} = \frac{{\left[ {{H_3}{O^ + }} \right] \times \left[ {{F^ - }} \right]}}{{[HF]}}\\3.5 \times 1{0^{ - 4}} = \frac{{\left( {1 \times 1{0^{ - 7}} + x} \right) \times x}}{{1 \times 1{0^{ - 2}} - x}}\\3.5 \times 1{0^{ - 6}} - 3.5 \times 1{0^{ - 4}}x = 1 \times 1{0^{ - 7}}x + {x^2}\\0 = {x^2} + 3.5 \times 1{0^{ - 4}}x - 3.5 \times 1{0^{ - 6}}\\x = 17.04 \times 1{0^{ - 4}}\\\left[ {{H_3}{O^ + }} \right] = 1 \times 1{0^{ - 7}} + 17.04 \times 1{0^{ - 4}} = 17.04 \times 1{0^{ - 4}}\end{array}\)

10

Step 10: 

\(\begin{array}{l} A curve for a series of solutions of HF ( vertical axis \left[ {{H_3}{O^ + }} \right]and horizontal axs [HF] \\(ionized and nonionized)) \end{array}\)

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Most popular questions from this chapter

Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

(a) \(HT{e^ - }\)(as a base)

(b) \({\left( {C{H_3}} \right)_3}N{H^ + }\)

(c) \(HAs{O_4}^{3 - }\)(as a base)

(d) \(HO_2^ - \)(as a base)

(e) \({C_6}{H_5}N{H_3}^ + \)

(f) \(HSO_3^ - \)(as a base)

Calculate the \(pH\)and the\(pOH\) of each of the following solutions at \(2{5^o}C\) for which the substances ionize completely:

(a)\(0.000259M HCl{O_4}\)

(b)\(0.21M NaOH\)

(c)\(0.000071M Ba{(OH)_2}\)

(d) \(2.5M KOH\)

Explain why equilibrium calculations are not necessary to determine ionic concentrations in solutions of certain strong electrolytes such as. \(NaOH\;and\;HCl.\;\) Under what conditions are equilibrium calculations necessary as part of the determination of the concentrations of all ions of some other strong electrolytes in solution?

What are the\(pH\;and\;pOH\) of a solution of \(2.0\)\(M\)\(HCl\), which ionizes completely?

Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution
See all solutions

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