3. A chemistry student makes up a 0.10 M solution of NaF(aq) and finds the pH to be 8.09.
a. Explain why the presence of F- in this solution affects the pH.
b. Write out the acid/base equilibrium equation for the reaction of F- with water.
c. Use the measured pH value to solve for the Kb of this reaction.
d. Calculate the product of your calculated Kb and the Ka of HF (found on the table) and explain the familiarity of this number.
I will do b and c and from that work you can do a and d.
b.
.........F^- + HOH ==> HF + OH^-
initial.0.1M...........0......0
c.
What pH value did you get? I need a number; let's call that y = OH and and you can substitute the value you measured it to be after converting from pH.
..........F^- HOH ==> HF + OH^-
initial..0.1..........0.....0
change...-y...........y.....y
equil....0.1-y.........y.....y
Kb for F^- = (HF)(OH^-)/(F^-)
You know HF from the measured pH.
You know OH from the measured pH.
You know F^- = 0.1-y
Solve for Kb for F^-
a. The presence of F- ions in the solution affects the pH because F- ions can react with water to form hydrofluoric acid (HF) and hydroxide ions (OH-). This reaction can introduce additional H+ (protons) or OH- ions into the solution, which can affect the overall concentration of H+ ions and thus the pH.
b. The acid/base equilibrium equation for the reaction of F- with water is:
F- + H2O ↔ HF + OH-
In this reaction, F- acts as a base and accepts a proton (H+) from water to form HF. The OH- ion is also formed as a result of the reaction.
c. To solve for the Kb of this reaction, we can use the fact that the concentration of OH- ions is related to the pH by the equation:
pOH = -log[OH-]
Since the pH of the solution is given as 8.09, we can calculate the pOH as:
pOH = 14 - pH
= 14 - 8.09
= 5.91
Next, we can convert the pOH back to the concentration of OH- ions:
[OH-] = 10^(-pOH)
= 10^(-5.91)
The Kb value can be calculated using the equation:
Kb = [HF][OH-]/[F-]
Since the concentration of F- ions is given as 0.10 M, and the concentration of OH- ions is determined as above, we can substitute these values into the equation to solve for Kb.
d. To calculate the product of Kb and Ka of HF, we need the value of Ka from the table. Unfortunately, the table is not provided in the question, so we cannot perform this calculation. However, if we had the value of Ka for HF, we could multiply it by the calculated Kb to obtain a value. The familiarity of this number would depend on the actual values used, so without the specific data, we cannot explain its familiarity.