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Difficulty: Very hardRelative Strength and Ionization of Acids and Bases

Two weak monobasic acids, HX\text{HX} (Ka=1.0×10−5 mol dm−3K_a = 1.0 \times 10^{-5}\text{ mol dm}^{-3}) and HY\text{HY} (Ka=4.0×10−5 mol dm−3K_a = 4.0 \times 10^{-5}\text{ mol dm}^{-3}), are prepared as aqueous solutions. Solution X contains 0.40 mol dm−30.40\text{ mol dm}^{-3} of HX\text{HX}, whereas Solution Y contains 0.10 mol dm−30.10\text{ mol dm}^{-3} of HY\text{HY}. Based on Ostwald's dilution law and the principles of acid ionization, which of the following statements correctly compares the two solutions?

  1. A
    Solution X has a higher hydrogen ion concentration than Solution Y because HX\text{HX} is four times more concentrated than HY\text{HY}.
  2. B
    Solution Y has a higher hydrogen ion concentration than Solution X because HY\text{HY} has a larger acid dissociation constant (KaK_a) than HX\text{HX}.
  3. Both solutions have the same hydrogen ion concentration of 2.0×10−3 mol dm−32.0 \times 10^{-3}\text{ mol dm}^{-3}, but HY\text{HY} has a four-fold greater degree of ionization than HX\text{HX}.Answer
  4. D
    Both solutions have the same degree of ionization, but Solution X contains a higher concentration of hydrogen ions.

Answer

Both solutions have the exact same hydrogen ion concentration of 2.0×10−3 mol dm−32.0 \times 10^{-3}\text{ mol dm}^{-3}, but HY\text{HY} has a four-fold greater degree of ionization than HX\text{HX}.
Evaluating [H+]=Ka⋅C[H^+] = \sqrt{K_a \cdot C} for both solutions gives [H+]X=1.0×10−5×0.40=2.0×10−3 mol dm−3[H^+]_X = \sqrt{1.0 \times 10^{-5} \times 0.40} = 2.0 \times 10^{-3}\text{ mol dm}^{-3} and [H+]Y=4.0×10−5×0.10=2.0×10−3 mol dm−3[H^+]_Y = \sqrt{4.0 \times 10^{-5} \times 0.10} = 2.0 \times 10^{-3}\text{ mol dm}^{-3}, showing both solutions have identical hydrogen ion concentrations. Calculating the degree of ionization α=Ka/C\alpha = \sqrt{K_a / C} yields αX=0.005\alpha_X = 0.005 (0.5%0.5\%) and αY=0.020\alpha_Y = 0.020 (2.0%2.0\%), demonstrating that HY\text{HY} is four times more ionized than HX\text{HX} in these conditions.

Step-by-Step Solution

1
Calculate [H+][H^+] for Solution X
[H+]X=Ka,X×CX=(1.0×10−5)×0.40=4.0×10−6=2.0×10−3 mol dm−3[H^+]_X = \sqrt{K_{a,X} \times C_X} = \sqrt{(1.0 \times 10^{-5}) \times 0.40} = \sqrt{4.0 \times 10^{-6}} = 2.0 \times 10^{-3}\text{ mol dm}^{-3}
For a weak monobasic acid, [H+]=Ka⋅C[H^+] = \sqrt{K_a \cdot C} derived from Ostwald's dilution law.
2
Calculate [H+][H^+] for Solution Y
[H+]Y=Ka,Y×CY=(4.0×10−5)×0.10=4.0×10−6=2.0×10−3 mol dm−3[H^+]_Y = \sqrt{K_{a,Y} \times C_Y} = \sqrt{(4.0 \times 10^{-5}) \times 0.10} = \sqrt{4.0 \times 10^{-6}} = 2.0 \times 10^{-3}\text{ mol dm}^{-3}
Applying the weak acid ionization expression to Solution Y.
3
Calculate the degree of ionization (\alpha) for both acids
αX=Ka,XCX=1.0×10−50.40=5.0×10−3=0.5%\alpha_X = \sqrt{\frac{K_{a,X}}{C_X}} = \sqrt{\frac{1.0 \times 10^{-5}}{0.40}} = 5.0 \times 10^{-3} = 0.5\%; αY=Ka,YCY=4.0×10−50.10=2.0×10−2=2.0%\alpha_Y = \sqrt{\frac{K_{a,Y}}{C_Y}} = \sqrt{\frac{4.0 \times 10^{-5}}{0.10}} = 2.0 \times 10^{-2} = 2.0\%
The degree of ionization is given by α=KaC\alpha = \sqrt{\frac{K_a}{C}}.
4
Compare the calculated parameters
[H+]X=[H+]Y=2.0×10−3 mol dm−3[H^+]_X = [H^+]_Y = 2.0 \times 10^{-3}\text{ mol dm}^{-3} and αYαX=2.0%0.5%=4\frac{\alpha_Y}{\alpha_X} = \frac{2.0\%}{0.5\%} = 4
Comparing [H+][H^+] and the ratio of degrees of ionization.

Key Concept

Relative Strength and Ionization of Acids and Bases
Estimated Time:2m 0s
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Relative Strength and Ionization of Acids and Bases

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