Finding the [OH-] and pH of a Weak Base Solution Find the OH- and pH of a 0.100 M NH3 solution.

Question

Finding the [OH[latex]^{-}[/latex]] and pH of a Weak Base Solution

Find the OH[latex]^{-}[/latex] and pH of a 0.100 M NH[latex]_{3}[/latex] solution.

Accepted Answer

1. Write the balanced equation for the ionization of water by the base and use it as a guide to prepare an ICE table showing the given concentration of the weak base as its initial concentration. Leave room in the table for the chang es in concentrations and for the equilibrium concentrations. (Note that you list the OH[latex]^{-}[/latex] concentration as approximately zero. Although a little OH[latex]^{-}[/latex] is present from the autoionization of water, this amount is negligibly small com pared to the amount of OH[latex]^{-}[/latex] formed by the base.)

[latex]NH_{3}(aq) + H_{2}O(l)\xrightleftharpoons[]{}NH_{4}^{+}(aq) + OH^{-}(aq)[/latex]

	[NH[latex]_{3}[/latex]]	[NH[latex_{4}^{+}[/latex]]	[OH[latex]^{-}[/latex]]
Initial	0.100	0.00	≈0.00
Change
Equil

2. Represent the change in the concentration of OH[latex]^{-}[/latex] with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x.

[latex]NH_{3}(aq) + H_{2}O(l)\xrightleftharpoons[]{}NH_{4}^{+}(aq) + OH^{-}(aq)[/latex]

	[NH[latex]_{3}[/latex]]	[NH[latex_{4}^{+}[/latex]]	[OH[latex]^{-}[/latex]]
Initial	0.100	0.00	≈0.00
Change	-x	+x	+x
Equil

3. Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

[latex]NH_{3}(aq) + H_{2}O(l)\xrightleftharpoons[]{}NH_{4}^{+}(aq) + OH^{-}(aq)[/latex]

	[NH[latex]_{3}[/latex]]	[NH[latex_{4}^{+}[/latex]]	[OH[latex]^{-}[/latex]]
Initial	0.100	0.00	≈0.00
Change	-x	+x	+x
Equil	0.100-x	x	x

4. Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the base ionization constant.
In many cases, you can make the approximation that x is small (as discussed in Chapter 14).
Substitute the value of the base ionization constant (from Table 15.9) into the K[latex]_{b}[/latex] expression and solve for x.
Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%).

[latex]K_{b} = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}[/latex]
=[latex]\frac{ x^{2}}{0.100 - \cancel{x}}[/latex] (x is small)
1.76× 10[latex]^{-5} = \frac{x^{2}}{0.100}[/latex]
[latex]\sqrt{1.76 imes 10^{-5}} = \sqrt{\frac{x^{2}}{0.100}}[/latex]
x = [latex]\sqrt{(0.100)(1.76 imes 10^{-5})}[/latex]
= 1.33 × 10[latex]^{-3}[/latex]
[latex]\frac{1.33 imes 10^{-3}}{0.100}[/latex]× 100% = 1.33%
Therefore, the approximation is valid.

5. Determine the OH- concentration from the calculated value of x.
Use the expression for K[latex]_{w}[/latex] to find[latex] [H_{3}O^{+}][/latex].
Substitute [latex][H_{3}O^{+}][/latex] into the pH equation to find pH.

[OH[latex]^{-}[/latex]] = 1.33 × 10[latex]^{-3}[/latex] M
[latex][H_{3}O^{+}][OH^{-}] = K_{w} = 1.00 imes 10^{-14}[/latex]
[latex][H_{3}O^{+}](1.33 imes 10^{-3}) = 1.00 imes 10^{-14}[/latex]
[latex][H_{3}O^{+}] = 7.52 imes 10^{-12}[/latex] M
pH = -log[ [latex]H_{3}O^{+}[/latex] ]
= -log(7.52 × 10[latex]^{-12}[/latex])
= 11.124

Table 15.9 Some Common Weak Bases
Weak Base	Ionization Reaction	K[latex]_{b}[/latex]	pK[latex]_{b}[/latex]
Carbonate ion (CO[latex]_{3}^{2-}[/latex])*	[latex]CO_{3}^{2-}(aq) + H_{2}O(l) \xrightleftharpoons[]{} HCO_{3}^{-}(aq) + OH^{-}(aq)[/latex]	1.8×10[latex]^{-4}[/latex]	3.74
Methylamine ([latex]CH_{3}NH_{2}[/latex])	[latex]CH_{3}NH_{2}(aq) + H_{2}O(l)\xrightleftharpoons[]{} CH_{3}NH_{3}^{+}(aq) + OH^{-}(aq)[/latex]	4.4×10[latex]^{-4}[/latex]	3.36
Ethylamine ([latex]C_{2}H_{5}NH_{2}[/latex])	[latex]C_{2}H_{5}NH_{2}(aq) + H_{2}O(l)\xrightleftharpoons[]{} C_{2}H_{5}NH_{3}^{+}(aq) + OH^{-}(aq)[/latex]	5.6×10[latex]^{-4}[/latex]	3.25
Ammonia (NH[latex]_{3}[/latex])	[latex]NH_{3}(aq) + H_{2}O(l)\xrightleftharpoons[]{} NH_{4}^{+}(aq) + OH^{-}(aq)[/latex]	1.76×10[latex]^{-5}[/latex]	3.754
Pyridine ([latex]C_{5}H_{5}N[/latex])	[latex]C_{5}H_{5}N(aq) + H_{2}O(l)\xrightleftharpoons[]{} C_{5}H_{5}NH^{+}(aq) + OH^{-}(aq)[/latex]	1.7×10[latex]^{-9}[/latex]	8.77
Bicarbonate ion (HCO[latex]_{3}^{-}[/latex])* (or hydrogen carbonate)	[latex]HCO_{3}^{-}(aq) + H_{2}O(l)\xrightleftharpoons[]{} H_{2}CO_{3}(aq) + OH^{-}(aq)[/latex]	1.7×10[latex]^{-9}[/latex]	8.77
Aniline ([latex]C_{6}H_{5}NH_{2}[/latex])	[latex]C_{6}H_{5}NH_{2}(aq) + H_{2}O(l)\xrightleftharpoons[]{} C_{6}H_{5}NH_{3}^{+}(aq) + OH^{-}(aq)[/latex]	3.9×10[latex]^{-10}[/latex]	9.41
*The carbonate and bicarbonate ions must occur with a positively charged ion such as Na[latex]^{+}[/latex] that serves to balance the charge but does not have any part in the ionization reaction. For example, it is the bicarbonate ion that makes sodium bicarbonate (NaHCO[latex]_{3}[/latex]) basic. We look more closely at ionic bases in Section 15.8.

Table 15.9 Some Common Weak Bases
Weak Base	Ionization Reaction	K $_{b}$	pK $_{b}$
Carbonate ion (CO $_{3}^{2-}$ )*	$CO_{3}^{2-}(aq) + H_{2}O(l) \xrightleftharpoons[]{} HCO_{3}^{-}(aq) + OH^{-}(aq)$	1.8×10 $^{-4}$	3.74
Methylamine ( $CH_{3}NH_{2}$ )	$CH_{3}NH_{2}(aq) + H_{2}O(l)\xrightleftharpoons[]{} CH_{3}NH_{3}^{+}(aq) + OH^{-}(aq)$	4.4×10 $^{-4}$	3.36
Ethylamine ( $C_{2}H_{5}NH_{2}$ )	$C_{2}H_{5}NH_{2}(aq) + H_{2}O(l)\xrightleftharpoons[]{} C_{2}H_{5}NH_{3}^{+}(aq) + OH^{-}(aq)$	5.6×10 $^{-4}$	3.25
Ammonia (NH $_{3}$ )	$NH_{3}(aq) + H_{2}O(l)\xrightleftharpoons[]{} NH_{4}^{+}(aq) + OH^{-}(aq)$	1.76×10 $^{-5}$	3.754
Pyridine ( $C_{5}H_{5}N$ )	$C_{5}H_{5}N(aq) + H_{2}O(l)\xrightleftharpoons[]{} C_{5}H_{5}NH^{+}(aq) + OH^{-}(aq)$	1.7×10 $^{-9}$	8.77
Bicarbonate ion (HCO $_{3}^{-}$ )* (or hydrogen carbonate)	$HCO_{3}^{-}(aq) + H_{2}O(l)\xrightleftharpoons[]{} H_{2}CO_{3}(aq) + OH^{-}(aq)$	1.7×10 $^{-9}$	8.77
Aniline ( $C_{6}H_{5}NH_{2}$ )	$C_{6}H_{5}NH_{2}(aq) + H_{2}O(l)\xrightleftharpoons[]{} C_{6}H_{5}NH_{3}^{+}(aq) + OH^{-}(aq)$	3.9×10 $^{-10}$	9.41
*The carbonate and bicarbonate ions must occur with a positively charged ion such as Na $^{+}$ that serves to balance the charge but does not have any part in the ionization reaction. For example, it is the bicarbonate ion that makes sodium bicarbonate (NaHCO $_{3}$ ) basic. We look more closely at ionic bases in Section 15.8.

Question 15.11: Finding the [OH-] and pH of a Weak Base Solution Find the OH...

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