Using data from Appendix D in the textbook, calculate [OH−] and pH for each of the following solutions.
Calculate the [OH] and the pH.
A mixture that is 5.2×10−2 M in NaC2H3O2 and 5.7×10−2 M in Ba(C2H3O2)2.
The correct answer and explanation is :
Final Answer:
- [OH⁻] = 9.60 × 10⁻⁶ M
- pH = 8.98
Explanation (300+ words):
To calculate the hydroxide ion concentration ([OH⁻]) and the pH of the given solution, we must understand the chemistry of the solutes involved:
The solution contains:
- Sodium acetate (NaC₂H₃O₂), a salt of a weak acid (acetic acid), which dissociates to give acetate ions (C₂H₃O₂⁻).
- Barium acetate [Ba(C₂H₃O₂)₂], which dissociates to give two acetate ions per formula unit.
First, we calculate the total acetate ion concentration contributed by both salts:
- Sodium acetate provides 0.052 M acetate ions.
- Barium acetate (0.057 M) provides 0.057 × 2 = 0.114 M acetate ions.
- Total acetate ion concentration = 0.052 + 0.114 = 0.166 M
The acetate ion is the conjugate base of acetic acid (CH₃COOH), a weak acid. It undergoes hydrolysis in water:
[
C₂H₃O₂⁻ + H₂O ⇌ HC₂H₃O₂ + OH⁻
]
This reaction produces hydroxide ions, making the solution basic.
We use the base dissociation constant (K_b) of the acetate ion to determine ([OH⁻]). (K_b) can be derived from the relation:
[
K_b = \frac{K_w}{K_a}
]
Where:
- (K_w = 1.0 × 10^{-14}) (ion-product of water),
- (K_a = 1.8 × 10^{-5}) (for acetic acid from Appendix D).
[
K_b = \frac{1.0 × 10^{-14}}{1.8 × 10^{-5}} ≈ 5.56 × 10^{-10}
]
Now apply the formula:
[
[OH⁻] ≈ \sqrt{K_b × [C₂H₃O₂⁻]} = \sqrt{5.56 × 10^{-10} × 0.166} ≈ 9.60 × 10^{-6} \text{ M}
]
Next, we find pOH:
[
pOH = -\log_{10}[OH⁻] ≈ -\log_{10}(9.60 × 10^{-6}) ≈ 5.02
]
Then, calculate pH:
[
pH = 14 – pOH ≈ 14 – 5.02 = 8.98
]
Thus, the solution is slightly basic due to the presence of the weak base (acetate ion).
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