The pH of the solution was 8.93 at 25.0 °C. Kw = 1.00 x 10^-14 M^2. Numbers with exponents like Kw must be input as -1.00e-14. {As usual, pay attention to significant figures.} Determine the following: The concentration of HF at equilibrium, the Kb of F^-, and the Ka of HF.
The Correct Answer and Explanation is:
To solve this problem, let’s break it down step by step:
Step 1: Understanding the Solution
We have a 0.350 mol/L solution of NaF. When NaF dissociates in water, it forms Na⁺ and F⁻ ions:
NaF→Na++F−\text{NaF} \rightarrow \text{Na}^+ + \text{F}^-NaF→Na++F−
The pH of the solution is given as 8.93. To find the pOH, we use the following relationship:pH+pOH=14\text{pH} + \text{pOH} = 14pH+pOH=14pOH=14−8.93=5.07\text{pOH} = 14 – 8.93 = 5.07pOH=14−8.93=5.07
The concentration of hydroxide ions (OH⁻) can be calculated using:[OH−]=10−pOH=10−5.07=8.51×10−6 M[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-5.07} = 8.51 \times 10^{-6} \, \text{M}[OH−]=10−pOH=10−5.07=8.51×10−6M
Step 2: F⁻ Hydrolysis
The fluoride ion (F⁻) is a weak base and will react with water to form HF and OH⁻ ions:F−+H2O⇌HF+OH−\text{F}^- + \text{H}_2\text{O} \rightleftharpoons \text{HF} + \text{OH}^-F−+H2O⇌HF+OH−
The concentration of OH⁻ ions at equilibrium is 8.51 × 10⁻⁶ M. The reaction begins with 0.350 M of F⁻ (since it comes directly from NaF dissociation).
Let’s use the equilibrium expression for the base dissociation constant (Kb) of F⁻:Kb=[HF][OH−][F−]\text{Kb} = \frac{[\text{HF}][\text{OH}^-]}{[\text{F}^-]}Kb=[F−][HF][OH−]
Step 3: Set Up the ICE Table
Let’s assume the change in the concentration of F⁻ is xxx, and the concentrations of HF and OH⁻ at equilibrium are both xxx. Initially, the concentration of F⁻ is 0.350 M.[F−][HF][OH−]Initial (M)0.35000Change (M)−xxxEquilibrium (M)0.350−xx8.51×10−6\begin{array}{|c|c|c|c|} \hline & [\text{F}^-] & [\text{HF}] & [\text{OH}^-] \\ \hline \text{Initial (M)} & 0.350 & 0 & 0 \\ \text{Change (M)} & -x & x & x \\ \text{Equilibrium (M)} & 0.350 – x & x & 8.51 \times 10^{-6} \\ \hline \end{array}Initial (M)Change (M)Equilibrium (M)[F−]0.350−x0.350−x[HF]0xx[OH−]0x8.51×10−6
Since the concentration of OH⁻ at equilibrium is 8.51 × 10⁻⁶ M, we can say that:x=8.51×10−6 Mx = 8.51 \times 10^{-6} \, \text{M}x=8.51×10−6M
Step 4: Solving for Kb of F⁻
Now we can substitute these values into the equilibrium expression for Kb:Kb=(8.51×10−6)(8.51×10−6)0.350−8.51×10−6\text{Kb} = \frac{(8.51 \times 10^{-6})(8.51 \times 10^{-6})}{0.350 – 8.51 \times 10^{-6}}Kb=0.350−8.51×10−6(8.51×10−6)(8.51×10−6)
Since 8.51×10−68.51 \times 10^{-6}8.51×10−6 is much smaller than 0.350, we can approximate:Kb=(8.51×10−6)20.350\text{Kb} = \frac{(8.51 \times 10^{-6})^2}{0.350}Kb=0.350(8.51×10−6)2Kb=7.24×10−110.350=2.07×10−10 M\text{Kb} = \frac{7.24 \times 10^{-11}}{0.350} = 2.07 \times 10^{-10} \, \text{M}Kb=0.3507.24×10−11=2.07×10−10M
Step 5: Finding Ka of HF
Next, we can find the Ka of HF using the relationship between Ka and Kb:Ka×Kb=Kw=1.00×10−14 M2\text{Ka} \times \text{Kb} = \text{Kw} = 1.00 \times 10^{-14} \, \text{M}^2Ka×Kb=Kw=1.00×10−14M2Ka=KwKb=1.00×10−142.07×10−10=4.83×10−5 M\text{Ka} = \frac{\text{Kw}}{\text{Kb}} = \frac{1.00 \times 10^{-14}}{2.07 \times 10^{-10}} = 4.83 \times 10^{-5} \, \text{M}Ka=KbKw=2.07×10−101.00×10−14=4.83×10−5M
Step 6: Concentration of HF at Equilibrium
Finally, the concentration of HF at equilibrium is equal to xxx, which is:[HF]=8.51×10−6 M[\text{HF}] = 8.51 \times 10^{-6} \, \text{M}[HF]=8.51×10−6M
Final Answers:
- The concentration of HF at equilibrium = 8.51×10−6 M8.51 \times 10^{-6} \, \text{M}8.51×10−6M
- The Kb of F⁻ = 2.07×10−10 M2.07 \times 10^{-10} \, \text{M}2.07×10−10M
- The Ka of HF = 4.83×10−5 M4.83 \times 10^{-5} \, \text{M}4.83×10−5M
These are the results based on the calculations.
