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Mobile App Chapter 14: Q114 E (page 837)
Draw a curve similar to that shown in Figure 14.23 for a series of solutions of\(N{H_3}\) . Plot [OH] on the vertical axis and the total concentration of \(N{H_3}\) (both ionized and nonionized \(N{H_3}\) molecules) on the horizontal axis. Let the total concentration of \(N{H_3}\) vary from \(1 \times 1{0^{ - 10}}M\) to \(1 \times 1{0^{ - 2}}M.\)
Short Answer
\(\begin{array}{*{20}{l}}{\left[ {{\rm{N}}{{\rm{H}}_3}} \right] = 1 \cdot {{10}^{ - 10}}{\rm{M}}/\left[ {{\rm{O}}{{\rm{H}}^ - }} \right] = 1.001 \cdot {{10}^{ - 7}}{\rm{M}}}\\{\left[ {{\rm{N}}{{\rm{H}}_3}} \right] = 1 \cdot {{10}^{ - 9}}{\rm{M}}/\left[ {{\rm{O}}{{\rm{H}}^ - }} \right] = 1.01 \cdot {{10}^{ - 7}}{\rm{M}}}\\{\left[ {{\rm{N}}{{\rm{H}}_3}} \right] = 1 \cdot {{10}^{ - 8}}{\rm{M}}/\left[ {O{H^ - }} \right] = 1.1 \cdot {{10}^{ - 7}}{\rm{M}}}\\{\left[ {{\rm{N}}{{\rm{H}}_3}} \right] = 1 \cdot {{10}^{ - 7}}{\rm{M}}/\left[ {{\rm{O}}{{\rm{H}}^ - }} \right] = 2 \cdot {{10}^{ - 7}}{\rm{M}}}\\{\left[ {{\rm{N}}{{\rm{H}}_3}} \right] = 1 \cdot {{10}^{ - 6}}{\rm{M}}/\left[ {O{H^ - }} \right] = 1.05 \cdot {{10}^{ - 6}}{\rm{M}}}\\{\left[ {{\rm{N}}{{\rm{H}}_3}} \right] = 1 \cdot {{10}^{ - 5}}{\rm{M}}/\left[ {O{H^ - }} \right] = 7.23 \cdot {{10}^{ - 6}}{\rm{M}}}\\{\left[ {{\rm{N}}{{\rm{H}}_3}} \right] = 1 \cdot {{10}^{ - 4}}{\rm{M}}/\left[ {{\rm{O}}{{\rm{H}}^ - }} \right] = 3.44 \cdot {{10}^{ - 5}}{\rm{M}}}\\{\left[ {{\rm{N}}{{\rm{H}}_3}} \right] = 1 \cdot {{10}^{ - 3}}{\rm{M}}/\left[ {{\rm{O}}{{\rm{H}}^ - }} \right] = 1.25 \cdot {{10}^{ - 4}}{\rm{M}}}\\{\left[ {{\rm{N}}{{\rm{H}}_3}} \right] = 1 \cdot {{10}^{ - 2}}{\rm{M}}/\left[ {O{H^ - }} \right] = 4.15 \cdot {{10}^{ - 4}}{\rm{M}}}\end{array}\)
Step by step solution
To find the concentration on \({\bf{N}}{{\bf{H}}_{\bf{3}}}{\bf{varies 1 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}{\bf{M}}\)
The reaction
-\({K_b}\)of\(N{H_3}\)is\(1.8 \times 1{0^{ - 5}}\)
- Concentration of\(N{H_3}\)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\(N{H_3}\)( vertical axis\(\left[ {O{H^ - }} \right]\)and horizontal axs\(\left[ {N{H_3}} \right]\)(ionized and nonionized) )
- Since the concentration of\(O{H^ - }\)ion in pure water is\(1 \times 1{0^{ - 7}}\), the initial concentration of\(O{H^ - }\)in a given solution will be\(1 \cdot 1{0^{ - 7}}.\)
\(\left[ {N{H_3}} \right] = 1 \times 1{0^{ - 10}}M\)
\(N{H_3}(aq) \to \) | \(\)\(N{H_4}^ + (aq) + \) | \(OH(aq)\) | |
Initial (M) | \({1.10^{ - 10}}\) | \(0\) | \({1.10^{ - 7}}\) |
Change (M) | -X | +X | +X |
Equilibrium (M) | \({1.10^{ - 10}} - X\) | X | \({1.10^{ - 7}} + X\) |
\(\begin{array}{l}{K_b} = \frac{{\left[ {{\rm{NH}} + {\rm{H}}_4^ + } \right] \cdot \left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}}{{\left[ {{\rm{N}}{{\rm{H}}_3}} \right]}}\\1.8 \cdot {10^{ - 5}} = \frac{{x \cdot \left( {1 \cdot {{10}^{ - 7}} + x} \right)}}{{1 \cdot {{10}^{ - 10}} - x}}\\1.8 \cdot {10^{ - 15}} - 1.8 \cdot {10^{ - 5}}x = 1 \cdot {10^{ - 7}}x + {x^2}\\0 = {x^2} + 1.81 \cdot {10^{ - 5}}x - 1.8 \cdot {10^{ - 15}}\\x = 1 \cdot {10^{ - 10}}\\\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] = 1 \cdot {10^{ - 7}} + 1 \cdot {10^{ - 10}} = 1.001 \cdot {10^{ - 7}}\end{array}\)
To find the concentration on \(\left[ {{\bf{N}}{{\bf{H}}_{\bf{3}}}} \right]{\bf{ varies 1 \times 1}}{{\bf{0}}^{{\bf{ - 9}}}}{\bf{M}}\)
\(\left[ {N{H_3}} \right] = 1 \times 1{0^{ - 9}}M\)
\(N{H_3}(aq) \to \) | \(\)\(N{H_4}^ + (aq) + \) | \(OH(aq)\) | |
Initial (M) | \({1.10^{ - 9}}\) | \(0\) | \({1.10^{ - 7}}\) |
Change (M) | -X | +X | +X |
Equilibrium (M) | \({1.10^{ - 9}} - X\) | X | \({1.10^{ - 7}} + X\) |
\(\begin{array}{l}{K_b} = \frac{{\left[ {N{\rm{ I }}_4^ + } \right] \cdot \left[ {O{1^ - }} \right]}}{{\left[ {{N_3}} \right]}}\\1.8 \cdot {10^{ - 5}} = \frac{{x \cdot \left( {1 \cdot {{10}^7} + x} \right)}}{{1 \cdot {{10}^{ - 9}} - x}}\\1.8 \cdot {10^{ - 14}} - 1.8 \cdot {10^{ - 5}}x = 1 \cdot {10^{ - 7}}x + {x^{\scriptstyle2\atop\scriptstyle}}\\0 = {x^2} + 1.81 \cdot {10^{ - 5}}x - 1.8 \cdot {10^{ - 14}}\\x = 1 \cdot {10^{ - 9}}\\\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] = 1 \cdot {10^{ - 7}} + 1 \cdot {10^{ - 9}} = 1.01 \cdot {10^{ - 7}}\end{array}\)
To find the concentration on \(\left[ {{\bf{N}}{{\bf{H}}_{\bf{3}}}} \right]{\bf{ varies 1 \times 1}}{{\bf{0}}^{{\bf{ - 8}}}}{\bf{ M}}\)
\(\left[ {N{H_3}} \right] = 1 \times 1{0^{ - 8}} M\)
\(N{H_3}(aq) \to \) | \(\)\(N{H_4}^ + (aq) + \) | \(OH(aq)\) | |
Initial (M) | \({1.10^{ - 8}}\) | \(0\) | \({1.10^{ - 7}}\) |
Change (M) | -X | +X | +X |
Equilibrium (M) | \({1.10^{ - 8}} - X\) | X | \({1.10^{ - 7}} + X\) |
\(\begin{array}{l}{K_b} = \frac{{\left[ {NH_1^ + } \right] \cdot \left[ {O{H^ - }} \right]}}{{\left[ {N{H_3}} \right]}}\\1.8 \cdot {10^{ - 5}} = \frac{{x \cdot \left( {1 \cdot {{10}^{ - 7}} + x} \right)}}{{1 \cdot {{10}^{ - 8}} - x}}\\1.8 \cdot {10^{ - 13}} - 1.8 \cdot {10^{ - 5}}x = 1 \cdot {10^{ - 7}}x + {x^2}\\0 = {x^2} + 1.81 \cdot {10^{ - 5}}x - 1.8 \cdot {10^{ - 13}}\\x = 1 \cdot {10^{ - 8}}\\\left[ {{H_3}{O^ + }} \right] = 1 \cdot {10^{ - 7}} + 1 \cdot {10^{ - 8}} = 1.1 \cdot {10^{ - 7}}\end{array}\)
To find the concentration on \(\left[ {{\bf{N}}{{\bf{H}}_{\bf{3}}}} \right]{\bf{ varies 1 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{M}}\)
\(\left[ {N{H_3}} \right] = 1 \times 1{0^{ - 7}}M\)
\(N{H_3}(aq) \to \) | \(\)\(N{H_4}^ + (aq) + \) | \(OH(aq)\) | |
Initial (M) | \({1.10^{ - 7}}\) | \(0\) | \({1.10^{ - 7}}\) |
Change (M) | -X | +X | +X |
Equilibrium (M) | \({1.10^{ - 7}} - X\) | X | \({1.10^{ - 7}} + X\) |
\(\begin{array}{l}{K_b} = \frac{{\left[ {{\rm{NH}}_1^ + } \right] \cdot \left[ {{\rm{OH}}{{\rm{H}}^ - }} \right]}}{{\left| {{\rm{N}}{{\rm{H}}_3}} \right|}}\\1.8 \cdot {10^{ - 5}} = \frac{{x \cdot \left( {1 \cdot {{10}^{ - 7}} + x} \right)}}{{1 \cdot {{10}^{ - 7}} - x}}\\1.8 \cdot {10^{ - 12}} - 1.8 \cdot {10^{ - 5}}x = 1 \cdot {10^{ - 7}}x + {x^2}\\0 = {x^2} + 1.81 \cdot {10^{ - 5}}x - 1.8 \cdot {10^{ - 12}}\\x = 1 \cdot {10^{ - 7}}\\\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] = 1 \cdot {10^{ - 7}} + 1 \cdot {10^{ - 7}} = 2 \cdot {10^{ - 7}}\end{array}\)
To find the concentration on \(\left[ {{\bf{N}}{{\bf{H}}_{\bf{3}}}} \right]{\bf{ varies 1 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{M}}\)
\(\left[ {N{H_3}} \right] = 1 \times 1{0^{ - 6}}M\)
\(N{H_3}(aq) \to \) | \(\)\(N{H_4}^ + (aq) + \) | \(OH(aq)\) | |
Initial (M) | \({1.10^{ - 6}}\) | \(0\) | \({1.10^{ - 7}}\) |
Change (M) | -X | +X | +X |
Equilibrium (M) | \({1.10^{ - 6}} - X\) | X | \({1.10^{ - 7}} + X\) |
\(\begin{array}{l}{K_b} = \frac{{\left[ {{\rm{NH}}_1^ + } \right] \cdot \left[ {{\rm{OH}}{{\rm{H}}^ - }} \right]}}{{\left| {{\rm{N}}{{\rm{H}}_3}} \right|}}\\1.8 \cdot {10^{ - 5}} = \frac{{x \cdot \left( {1 \cdot {{10}^{ - 7}} + x} \right)}}{{1 \cdot {{10}^{ - 6}} - x}}\\1.8 \cdot {10^{ - 11}} - 1.8 \cdot {10^{ - 5}}x = 1 \cdot {10^{ - 7}}x + {x^2}\\0 = {x^2} + 1.81 \cdot {10^{ - 5}}x - 1.8 \cdot {10^{ - 11}}\\x = 0.95 \cdot {10^{ - 6}}\\\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] = 1 \cdot {10^{ - 7}} + 0.95 \cdot {10^{ - 6}} = 1.05 \cdot {10^{ - 6}}\end{array}\)
To find the concentration on \(\left[ {{\bf{N}}{{\bf{H}}_{\bf{3}}}} \right]{\rm{ }}{\bf{varies}}{\rm{ }}{\bf{1 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}{\bf{M}}\)
\(\left[ {N{H_3}} \right] = 1 \times 1{0^{ - 5}}M\)
\(N{H_3}(aq) \to \) | \(\)\(N{H_4}^ + (aq) + \) | \(OH(aq)\) | |
Initial (M) | \({1.10^{ - 5}}\) | \(0\) | \({1.10^{ - 7}}\) |
Change (M) | -X | +X | +X |
Equilibrium (M) | \({1.10^{ - 5}} - X\) | X | \({1.10^{ - 7}} + X\) |
\(\begin{array}{l}{K_{\dot b}} = \frac{{\left[ {{\rm{NH}}_1^ + } \right] \cdot [{\rm{OH}} - ]}}{{\left. {\mid {\rm{N}}{{\rm{H}}_3}} \right]}}\\1.8 \cdot {10^{ - 5}} = \frac{{x \cdot \left( {1 \cdot {{10}^{ - 7}} + x} \right)}}{{1 \cdot {{10}^{ - 5}} - x}}\\1.8 \cdot {10^{ - 10}} - 1.8 \cdot {10^{ - 5}}x = 1 \cdot {10^{ - 7}}x + {x^2}\\0 = {x^2} + 1.81 \cdot {10^{ - 5}}x - 1.8 \cdot {10^{ - 10}}\\x = 7.13 \cdot {10^{ - 6}}\\\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] = 1 \cdot {10^{ - 7}} + 7.13 \cdot {10^{ - 6}} = 7.23 \cdot {10^{ - 6}}\end{array}\)
To find the concentration on \(\left[ {{\bf{N}}{{\bf{H}}_{\bf{3}}}} \right]{\bf{ varies 1 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{M}}\)
\(\left[ {N{H_3}} \right] = 1 \times 1{0^{ - 4}}M\)
\(N{H_3}(aq) \to \) | \(\)\(N{H_4}^ + (aq) + \) | \(OH(aq)\) | |
Initial (M) | \({1.10^{ - 4}}\) | \(0\) | \({1.10^{ - 7}}\) |
Change (M) | -X | +X | +X |
Equilibrium (M) | \({1.10^{ - 4}} - X\) | X | \({1.10^{ - 7}} + X\) |
\(\begin{array}{l}{K_b} = \frac{{\left[ {NH_1^ + } \right] \cdot \left[ {O{H^ - }} \right]}}{{\left[ {{\rm{N}}{{\rm{H}}_3}} \right]}}\\1.8 \cdot {10^{ - 5}} = \frac{{x \cdot \left( {1 \cdot {{10}^{ - 7}} + x} \right)}}{{1 \cdot {{10}^{ - 4}} - x}}\\1.8 \cdot {10^{ - 9}} - 1.8 \cdot {10^{ - 5}}x = 1 \cdot {10^{ - 7}}x + {x^2}\\0 = {x^2} + 1.81 \cdot {10^{ - 5}}x - 1.8 \cdot {10^{ - 9}}\\x = 3.43 \cdot {10^{ - 5}}\\\left[ {{H_3}{O^ + }} \right] = 1 \cdot {10^{ - 7}} + 3.43 \cdot {10^{ - 5}} = 3.44 \cdot {10^{ - 5}}\end{array}\)
To find the concentration on \(\left[ {{\bf{N}}{{\bf{H}}_{\bf{3}}}} \right]{\bf{ varies 1 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}{\bf{M}}\)
\(\left[ {N{H_3}} \right] = 1 \times 1{0^{ - 3}}M\)
\(N{H_3}(aq) \to \) | \(\)\(N{H_4}^ + (aq) + \) | \(OH(aq)\) | |
Initial (M) | \({1.10^{ - 3}}\) | \(0\) | \({1.10^{ - 7}}\) |
Change (M) | -X | +X | +X |
Equilibrium (M) | \({1.10^{ - 3}} - X\) | X | \({1.10^{ - 7}} + X\) |
\(\begin{array}{l}{K_b} = \frac{{\left[ {NH_1^ + } \right] \cdot \left[ {O{H^ - }} \right]}}{{\left[ {N{H_3}} \right]}}\\1.8 \cdot {10^{ - 5}} = \frac{{x \cdot \left( {1 \cdot {{10}^{ - 7}} + x} \right)}}{{1 \cdot {{10}^{ - 3}} - x}}\\1.8 \cdot {10^{ - 8}} - 1.8 \cdot {10^{ - 5}}x = 1 \cdot {10^{ - 7}}x + {x^2}\\0 = {x^2} + 1.81 \cdot {10^{ - 5}}x - 1.8 \cdot {10^{ - 8}}\\x = 1.25 \cdot {10^{ - 4}}\\\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] = 1 \cdot {10^{ - 7}} + 1.25 \cdot {10^{ - 4}} = 1.25 \cdot {10^{ - 4}}\end{array}\)
To find the concentration on \(\left[ {{\bf{N}}{{\bf{H}}_{\bf{3}}}} \right]{\bf{ varies 1 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}{\bf{M}}\)
\(\left[ {N{H_3}} \right] = 1 \times 1{0^{ - 2}}M\)
\(N{H_3}(aq) \to \) | \(\)\(N{H_4}^ + (aq) + \) | \(OH(aq)\) | |
Initial (M) | \({1.10^{ - 2}}\) | \(0\) | \({1.10^{ - 7}}\) |
Change (M) | -X | +X | +X |
Equilibrium (M) | \({1.10^{ - 2}} - X\) | X | \({1.10^{ - 7}} + X\) |
\(\begin{array}{l}{K_b} = \frac{{\left[ {NH_4^ + } \right] \cdot \left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}}{{\left. {\mid {\rm{N}}{{\rm{H}}_3}} \right]}}\\1.8 \cdot {10^{ - 5}} = \frac{{x \cdot \left( {1 \cdot {{10}^{ - 7}} + x} \right)}}{{1 \cdot {{10}^{ - 2}} - x}}\\1.8 \cdot {10^{ - 7}} - 1.8 \cdot {10^{ - 5}}x = 1 \cdot {10^{ - 7}}x + {x^2}\\0 = {x^2} + 1.81 \cdot {10^{ - 5}}x - 1.8 \cdot {10^{ - 7}}\\x = 4.15 \cdot {10^{ - 4}}\\\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] = 1 \cdot {10^{ - 7}} + 4.15 \cdot {10^{ - 4}} = 4.15 \cdot {10^{ - 4}}\end{array}\)
Step 10:
\(\begin{array}{l}{\rm{ A curve for a series of solutions of N}}{{\rm{H}}_3}{\rm{ ( vertical axis }}\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]{\rm{and horizontal axs }}\\\left[ {{\rm{N}}{{\rm{H}}_3}} \right]{\rm{ (ionized and nonionized)) }}\end{array}\)
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