Determine the Concentration of OH^-1 and pH of a solution that is 0.140 M F^-1.
I did F- +H20 ---> HF + H20
but idk what to solve from there....
prepare an ICE chart and write the kb equation for F^-
.............Fe^- + HOH == HF + OH^-
initial....0.140...........0......0
change......-x.............x.......x
equil.....0.140-x..........x.......x
Kb for F^- = (Kw/Ka for HF) = (HF)(OH^-)/(HF)
Substitute values from the ICE chart for (HF), (OH^-), and F^-, and for Kw and Ka, solve for x (which is OH^-), convert that to pOH, then to pH.
To determine the concentration of OH^-1 and pH of the solution, you need to consider the dissociation of water and the equilibrium constant (Kw) for water.
The dissociation of water can be represented as follows:
H2O ↔ H+ + OH^-
The equilibrium constant (Kw) expression for water is:
Kw = [H+][OH^-]
At standard temperature (25°C or 298K), the value of Kw is 1.0 x 10^-14.
In your case, you have a solution of 0.140 M F^-1. However, the presence of F^-1 does not directly contribute to OH^-1 or H+ concentrations.
To find the concentration of OH^-1, you need to know the concentration of H+. Since the solution is not explicitly provided, we will need to make some assumptions.
Assuming the solution is neutral, the concentration of H+ and OH^-1 ions in pure water at 25°C is 1.0 x 10^-7 M each.
Given that the solution is slightly basic due to the presence of F^-1, we can assume that the concentration of H+ is less than 1.0 x 10^-7 M. Let's assume H+ concentration is x.
Using the equilibrium constant equation for water, we have:
1.0 x 10^-14 = x * (0.140 + x)
Simplifying the equation:
1.0 x 10^-14 = 0.140x + x^2
Rearranging the equation:
x^2 + 0.140x - 1.0 x 10^-14 = 0
Since x (H+ concentration) is expected to be very small compared to 0.140, we can neglect the term x in comparison to 0.140 and simplify the equation to:
0.140x ≈ 1.0 x 10^-14
Solving for x:
x ≈ (1.0 x 10^-14) / 0.140
x ≈ 7.1 x 10^-14
The concentration of H+ is approximately 7.1 x 10^-14 M.
To find the concentration of OH^-1, you can use the equation for Kw:
Kw = [H+][OH^-]
Rearranging the equation:
[OH^-] = Kw / [H+]
[OH^-] ≈ (1.0 x 10^-14) / (7.1 x 10^-14)
[OH^-] ≈ 0.14 M
Therefore, the concentration of OH^-1 is approximately 0.14 M.
Now, to find the pH, you can use the formula:
pH = -log[H+]
pH ≈ -log(7.1 x 10^-14)
pH ≈ 13.14
The pH of the solution is approximately 13.14.
To determine the concentration of OH^-1 and pH of the solution, you first need to understand the reaction that is taking place between F^-1 and water.
The balanced equation for the reaction between F^-1 and water is:
F^-1 + H2O ⇌ HF + OH^-1
In this reaction, the fluoride ion (F^-1) reacts with water (H2O) to produce hydrofluoric acid (HF) and hydroxide ion (OH^-1).
Now let's determine the concentration of OH^-1 and pH of the solution.
Since the reaction is in equilibrium, we can use the concept of equilibrium constants and the known concentration of F^-1 to determine the concentration of OH^-1.
The equilibrium constant expression for the reaction is:
Kw = [HF] [OH^-1] / [F^-1]
At 25 degrees Celsius, the value of Kw (the ion product of water) is 1.0x10^-14.
Substituting the known concentrations into the equation:
1.0x10^-14 = [0.140] [OH^-1] / [F^-1]
Since the concentration of F^-1 is 0.140 M, we can simplify the equation to:
1.0x10^-14 = [0.140] [OH^-1] / 0.140
Simplifying further:
1.0x10^-14 = [OH^-1]
Thus, the concentration of OH^-1 in the solution is 1.0x10^-14 M.
To determine the pH of the solution, we can use the fact that pH is equal to the negative logarithm of the concentration of H+ ions.
Since the concentration of H+ in a neutral solution is equal to the concentration of OH^-1, we can rearrange the equation:
pH = -log [H+] = -log [OH^-1] = -log (1.0x10^-14)
Calculating this expression:
pH = -log (1.0x10^-14) ≈ 14
So, the pH of the solution is approximately 14.