Calculate the pH of a 0.053 M (C2H5)2NH solution (Kb = 1.3 10-3).
Oh, looks like we have an equation to calculate the pH! Now let's see... (C2H5)2NH is definitely a tongue twister, but let's not get too tongue-tied here. The compound (C2H5)2NH is actually a weak base, and Kb is the base dissociation constant. So, to Calculate the pH of a 0.053 M (C2H5)2NH solution, we need to keep calm and crunch some numbers.
Now, Kb is given as 1.3 * 10^-3. Are you ready for the fun part? Let's calculate it step by step. First, we need to find the concentration of OH- ions formed by the (C2H5)2NH. For that, we can use the Kb expression:
Kb = [OH-]^2 / [C2H5)2NH]
Since we're given the concentration of (C2H5)2NH (0.053 M), we'll be using that here. Let's say the concentration of OH- ions is x. So, Kb = (x^2) / (0.053).
Now, we can solve for x by rearranging the equation:
x^2 = Kb * (C2H5)2NH
x^2 = (1.3 * 10^-3) * (0.053)
x = sqrt((1.3 * 10^-3) * (0.053))
Okay, this is the serious part. Let's do some quick calculations and get the result.
Calculating... calculating... drumroll, please... and we have a result! Based on my calculations, the concentration of OH- ions comes out to be approximately 0.013 M.
Now, pH is calculated by subtracting the negative logarithm (base 10) of the concentration of H+ ions from 14. Since we have the concentration of OH- ions, we can relate it to the concentration of H+ ions through water's autoprotolysis equation:
[H+] * [OH-] = 1 * 10^-14.
Since we know the concentration of OH- ions (0.013 M), we can use this equation to find the concentration of H+ ions. After solving, we find it to be approximately 7.7 * 10^-13 M.
So, let me put on my clown nose and crunch some more numbers. Calculating... calculating... and voila! The pH of the 0.053 M (C2H5)2NH solution is approximately 12.11.
Please note that my calculations have been done with a sense of humor, so take them with a grain of salt.
To calculate the pH of a (C2H5)2NH solution, we need to find the concentration of hydroxide ions (OH-) first and then convert it to pH.
Step 1: Write the balanced equation for the reaction of (C2H5)2NH with water:
(C2H5)2NH + H2O ⇌ (C2H5)NH2OH- + OH-
Step 2: Write the Kb expression:
Kb = ([C2H5NH2OH-] * [OH-]) / [C2H5)2NH)
Step 3: Set up the ICE (initial, change, equilibrium) table:
(C2H5)2NH + H2O ⇌ (C2H5)NH2OH- + OH-
I 0.053 M 0 M 0 M
C -x +x +x
E 0.053 - x x x
Step 4: Substitute the equilibrium concentrations into the Kb expression:
1.3 × 10^-3 = x^2 / (0.053 - x)
Step 5: Simplify the equation:
1.3 × 10^-3 = x^2 / 0.053
Step 6: Rearrange the equation to solve for x:
x^2 = (1.3 × 10^-3) * 0.053
x^2 = 6.89 × 10^-5
x ≈ 8.30 × 10^-3 M
Step 7: Calculate the pOH:
pOH = -log10(OH-) = -log10(8.30 × 10^-3) ≈ 2.08
Step 8: Calculate the pH:
pH = 14 - pOH = 14 - 2.08 ≈ 11.92
The pH of a 0.053 M (C2H5)2NH solution with Kb = 1.3 × 10^-3 is approximately 11.92.
To calculate the pH of a solution, we need to determine the concentration of hydrogen ions (H+) in the solution. In this case, we are given the concentration of a base, (C2H5)2NH, and the value of its base dissociation constant, Kb.
To start, let's write the balanced chemical equation for the dissociation of (C2H5)2NH in water:
(C2H5)2NH + H2O ⇌ (C2H5)2NH2+ + OH-
Since we know the base dissociation constant, Kb, we can set up an equilibrium expression for this equation:
Kb = [C2H5)2NH2+] [OH-] / [(C2H5)2NH]
We are interested in finding the concentration of OH-. The concentration of OH- can be calculated by solving this equilibrium expression for OH-:
[OH-] = (Kb * [(C2H5)2NH]) / [(C2H5)2NH2+]
Given that (C2H5)2NH has a concentration of 0.053 M, and Kb = 1.3 * 10^-3, we can substitute these values into the equation:
[OH-] = (1.3 * 10^-3 * 0.053) / 0.053
Simplifying the equation:
[OH-] = 1.3 * 10^-3 M
Since the solution is basic, the concentration of H+ can be calculated by using the equation:
[H+] = Kw / [OH-]
where Kw is the ion product constant of water, which is equal to 1.0 * 10^-14 at 25°C.
[H+] = (1.0 * 10^-14) / (1.3 * 10^-3)
Simplifying the equation:
[H+] = 7.7 * 10^-12 M
Finally, we can calculate the pH of the solution using the equation:
pH = -log[H+]
pH = -log(7.7 * 10^-12)
Using logarithm properties, we get:
pH = 11.11
Therefore, the pH of the 0.053 M (C2H5)2NH solution is approximately 11.11.
2.13
..........B2NH + HOH ==> B2NH2^+ + OH^-
initial....0.053..........0.........0
change....-x..............x.........x
equil.....0.053-x.........x..........x
Kb = (B2NH^+)(OH^-)/(B2BH)
Substitute into Kb expression and solve for x (which is OH^-). Convert to pH.