Determining the pH of a Solution Containing an Anion Acting as a Base

Find the pH of a 0.100 M NaCHO $_{2}$ solution. The salt completely dissociates into Na $^{+}$ (aq) and CHO $_{2}^{-}$ (aq), and the Na $^{+}$ ion has no acid or base properties.

Question

Determining the pH of a Solution Containing an Anion Acting as a Base

Find the pH of a 0.100 M NaCHO[latex]_{2}[/latex] solution. The salt completely dissociates into Na[latex]^{+}[/latex](aq) and CHO[latex]_{2}^{-}[/latex](aq), and the Na[latex]^{+}[/latex] ion has no acid or base properties.

Accepted Answer

1. Since the Na+ ion does not have any acid or base properties, you can ignore it. Write the balanced equation for the ioniza tion of water by the basic anion and use it as a guide to pre pare an ICE table showing the given concentration of the weak base as its initial concentration.

[latex]CHO_{2}^{-}(aq) + H_{2}O(l) \xrightleftharpoons[]{} HCHO_{2}(aq) + OH^{-}(aq)[/latex]

	[CHO[latex]_{2}[/latex]]	[HCHO[latex]_{2}[/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]CHO_{2}^{-}(aq) + H_{2}O(l) \xrightleftharpoons[]{} HCHO_{2}(aq) + OH^{-}(aq)[/latex]

	[CHO[latex]_{2}[/latex]]	[HCHO[latex]_{2}[/latex]]	[OH[latex]^{-}[/latex]]
Initial	0.100	0.00	≈0.00
Change	-x	+x	+x
Equil

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

[latex]CHO_{2}^{-}(aq) + H_{2}O(l) \xrightleftharpoons[]{} HCHO_{2}(aq) + OH^{-}(aq)[/latex]

	[CHO[latex]_{2}[/latex]]	[HCHO[latex]_{2}[/latex]]	[OH[latex]^{-}[/latex]]
Initial	0.100	0.00	≈0.00
Change	-x	+x	+x
Equil	0.100-x	x	x

4. Find K[latex]_{b}[/latex] from K[latex]_{a}[/latex] (for the conjugate acid which you can find in Table 15.5).
Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for K[latex]_{b}[/latex]. In many cases, you can make the approximation that x is small (as discussed in Chapter 14).
Substitute the value of Kb into the K[latex]_{b}[/latex] expression and solve for x.
Confirm that the x is small approximation is valid by calculat-ing 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_{a} imes K_{b} = K_{w}[/latex]
K[latex]_{b} = \frac{K_{w}}{K_{a}}=\frac{ 1.0 imes 10^{-14}}{1.8 imes 10^{-4}} = 5.6 imes 10^{-11}[/latex]
K[latex]_{b} = \frac{[HCHO_{2}][OH^{-}]}{[CHO_{2}^{-}]}[/latex]
=[latex]\frac{ x^{2}}{0.100 - \cancel{x}}[/latex]
5.6× 10[latex]^{-11} = \frac{x^{2}}{0.100}[/latex]
x = 2.4 × 10[latex]^{-6}[/latex]
[latex]\frac{2.4 imes 10^{-6}}{0.100}[/latex]× 100% = 0.0024%
Therefore, the approximation is valid.

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

[OH[latex]^{-}[/latex]] = 2.4 × 10[latex]^{-6}[/latex] M
[latex][H_{3}O^{+}][OH^{-}] = K_{w} = 1.0 imes 10^{-14}[/latex]
[latex][H_{3}O^{+}](2.4 imes 10^{-6}) = 1.0 imes 10^{-14}[/latex]
[latex][H_{3}O^{+}] = 4.2 imes 10^{-9}[/latex] M
pH = -log [ [latex]H_{3}O^{+}[/latex] ]
= -log (4.2 × 10[latex]^{-9}[/latex])
= 8.38

TABLE 15.5 Acid Ionization Constants (K[latex]_{a}[/latex]) for Some Monoprotic Weak Acids at 25 °C
Acid	Formula	Structural Formula	Ionization Reaction	K[latex]_{a}[/latex]	pK*[latex]_{a}[/latex]
Chlorous acid	HCIO[latex]_{2}[/latex]	H—O—Cl=O	[latex]HClO_{2}(aq)+H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq)+ClO_{2}^{-}(aq)[/latex]	1.1×10[latex]^{-2}[/latex]	1.96
Nitrous acid	HNO[latex]_{2}[/latex]	H—O—N=O	[latex]HNO_{2}(aq)+ H_{2}O(l) \xrightleftharpoons[]{} H_{3}O^{+}(aq) + NO_{2}^{–}(aq)[/latex]	4.6×10[latex]^{-4}[/latex]	3.34
Hydrouoric acid	HF	H—F	[latex]HF(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + F^{–}(aq)[/latex]	3.5×10[latex]^{-4}[/latex]	3.46
Formic acid	HCHO[latex]_{2}[/latex]	[latex]H—O—\overset{O}{\underset{C}{\parallel } } —H[/latex]	[latex]HCHO_{2}(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O+(aq) + CHO_{2}^{–}(aq)[/latex]	1.8×10[latex]^{-4}[/latex]	3.74
Benzoic acid	[latex]HC_{7}H_{5}O_{2}[/latex]		[latex]HC_{7}H_{5}O_{2}(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + C_{7}H_{5}O_{2}^{–}[/latex](aq)	6.5×10[latex]^{-5}[/latex]	4.19
Acetic acid	[latex]HC_{2}H_{3}O_{2}[/latex]	[latex]H—O\overset{O}{\underset{C}{\parallel } } —CH_{3}[/latex]	[latex]HC_{2}H_{3}O_{2}(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + C_{2}H_{3}O_{2}^{–}(aq)[/latex]	1.8×10[latex]^{-5}[/latex]	4.74
Hypochlorous acid	HCIO[latex]_{2}[/latex]	H—O—CI	[latex]HClO(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + ClO^{–}(aq)[/latex]	2.9×10[latex]^{-8}[/latex]	7.54
Hydrocyanic acid	HCN	H—C≡N	[latex]HCN(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq)+CN^{–}(aq)[/latex]	4.9×10[latex]^{-10}[/latex]	9.31
Phenol	[latex]HC_{6}H_{5}O[/latex]		[latex]HC_{6}H_{5}O(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq)+C_{6}H_{5}O^{–}(aq)[/latex]	1.3×10[latex]^{-10}[/latex]	9.89
*pK[latex]_{a}[/latex] = –logK[latex]_{a}[/latex] (See the discussion at the end of Section 15.5)

TABLE 15.5 Acid Ionization Constants (K $_{a}$ ) for Some Monoprotic Weak Acids at 25 °C
Acid	Formula	Structural Formula	Ionization Reaction	K $_{a}$	pK* $_{a}$
Chlorous acid	HCIO $_{2}$	H—O—Cl=O	$HClO_{2}(aq)+H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq)+ClO_{2}^{-}(aq)$	1.1×10 $^{-2}$	1.96
Nitrous acid	HNO $_{2}$	H—O—N=O	$HNO_{2}(aq)+ H_{2}O(l) \xrightleftharpoons[]{} H_{3}O^{+}(aq) + NO_{2}^{–}(aq)$	4.6×10 $^{-4}$	3.34
Hydrouoric acid	HF	H—F	$HF(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + F^{–}(aq)$	3.5×10 $^{-4}$	3.46
Formic acid	HCHO $_{2}$	$H—O—\overset{O}{\underset{C}{\parallel } } —H$	$HCHO_{2}(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O+(aq) + CHO_{2}^{–}(aq)$	1.8×10 $^{-4}$	3.74
Benzoic acid	$HC_{7}H_{5}O_{2}$		$HC_{7}H_{5}O_{2}(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + C_{7}H_{5}O_{2}^{–}$ (aq)	6.5×10 $^{-5}$	4.19
Acetic acid	$HC_{2}H_{3}O_{2}$	$H—O\overset{O}{\underset{C}{\parallel } } —CH_{3}$	$HC_{2}H_{3}O_{2}(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + C_{2}H_{3}O_{2}^{–}(aq)$	1.8×10 $^{-5}$	4.74
Hypochlorous acid	HCIO $_{2}$	H—O—CI	$HClO(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + ClO^{–}(aq)$	2.9×10 $^{-8}$	7.54
Hydrocyanic acid	HCN	H—C≡N	$HCN(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq)+CN^{–}(aq)$	4.9×10 $^{-10}$	9.31
Phenol	$HC_{6}H_{5}O$		$HC_{6}H_{5}O(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq)+C_{6}H_{5}O^{–}(aq)$	1.3×10 $^{-10}$	9.89
*pK $_{a}$ = –logK $_{a}$ (See the discussion at the end of Section 15.5)

Question 15.13: Determining the pH of a Solution Containing an Anion Acting ...

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