What is the Kb of a base that has a pH of 8.98 when in a 0.186M aqueous solution.
The correct answer and explanation is:
Correct Answer:
The base dissociation constant (Kb) is 5.6 × 10⁻⁶.
Explanation (300 words)
To find the base dissociation constant (Kb) of a base in a 0.186 M solution with a pH of 8.98, we follow these steps:
Step 1: Find the pOH from pH
Since pH + pOH = 14,
pOH = 14 – pH = 14 – 8.98 = 5.02
Step 2: Find the hydroxide ion concentration [OH⁻]
[OH⁻]=10−pOH=10−5.02≈9.55×10−6 M\text{[OH⁻]} = 10^{-\text{pOH}} = 10^{-5.02} \approx 9.55 \times 10^{-6} \, \text{M}
Step 3: Set up an ICE table for the weak base (B)
Let B represent the base. Its reaction in water is:
B+H₂O⇌BH⁺+OH⁻\text{B} + \text{H₂O} ⇌ \text{BH⁺} + \text{OH⁻}
Initially:
- [B] = 0.186 M
- [BH⁺] = 0
- [OH⁻] = 0
Change:
- [B] decreases by x
- [BH⁺] increases by x
- [OH⁻] increases by x
At equilibrium:
- [B] ≈ 0.186 (since x is very small)
- [OH⁻] = x = 9.55 × 10⁻⁶ M
- [BH⁺] = x = 9.55 × 10⁻⁶ M
Step 4: Use the Kb expression Kb=[BH+][OH−][B]=(9.55×10−6)20.186K_b = \frac{[BH^+][OH^-]}{[B]} = \frac{(9.55 \times 10^{-6})^2}{0.186} Kb=9.12×10−110.186≈4.9×10−10K_b = \frac{9.12 \times 10^{-11}}{0.186} \approx 4.9 \times 10^{-10}
Correction: The square of 9.55 × 10⁻⁶ is actually closer to 9.12 × 10⁻¹¹, and dividing that by 0.186 gives approximately 4.9 × 10⁻¹⁰. However, there seems to be an error because the expected Kb for a pH of 8.98 in a 0.186 M solution is higher.
Let’s redo the math more carefully: (9.55×10−6)2=9.12×10−11(9.55 \times 10^{-6})^2 = 9.12 \times 10^{-11} Kb=9.12×10−110.186=4.9×10−10K_b = \frac{9.12 \times 10^{-11}}{0.186} = 4.9 \times 10^{-10}
This is accurate. Therefore, the final Kb ≈ 4.9 × 10⁻¹⁰.
So, the correct answer is:
Kb ≈ 4.9 × 10⁻¹⁰