If the Kb of a weak base is 1.5 × 10^-6, what is the pH of a 0.25 M solution of this base?
I found that I need to use the quadratic equation but I'm having difficulties because I come up with taking the square root by zero and that isn't right. Please help!
a=1, b=1.5x10^-6,
c=(1.5x10^-6(.25))=3.75x10^-7
Figured it out, it is 11. :)
You don't need to solve a quadratic.
1.5E-6 = (x)(x)/(0.25-x)
0.25-x = essentially 0.25 since x is very small.
x = 6.12E-4 = (OH^-)
Then pOH = ? and
pH + pOH = pKw = 14.
Solve for pH. something like 11 or so.
To find the pH of a weak base solution with a known Kb value, you can use the following steps:
Step 1: Write the balanced chemical equation for the dissociation of the weak base. Let's assume the weak base is represented as BH. Its dissociation in water can be written as:
BH (aq) + H2O (l) ⇌ B- (aq) + H3O+ (aq)
Step 2: Write the expression for the base dissociation constant (Kb) using the equilibrium concentrations of the products and reactants:
Kb = [B-][H3O+] / [BH]
Step 3: Since the concentration of the weak base is given as 0.25 M, we can assume that the concentration of B- at equilibrium is also 0.25 M, as the base fully dissociates.
Step 4: Let x be the concentration of H3O+ at equilibrium. As x is very small compared to 0.25 M, the change in concentration of BH due to the production of x can be approximated as (0.25 - x).
Step 5: Substitute the known values into the Kb expression:
1.5 × 10^-6 = (0.25 - x)(x) / (0.25)
Step 6: Rearrange the equation to solve for x:
1.5 × 10^-6 = (0.25 - x)x / (0.25)
1.5 × 10^-6 = (0.25x - x^2) / (0.25)
Multiply both sides of the equation by 0.25 to get rid of the denominator:
0.25 * 1.5 × 10^-6 = 0.25(0.25x - x^2) / (0.25)
0.375 × 10^-6 = 0.25x - x^2
Step 7: Rearrange the equation to form a quadratic equation:
x^2 - 0.25x + 0.375 × 10^-6 = 0
Now, let's solve the quadratic equation using the quadratic formula:
a = 1
b = -0.25
c = 0.375 × 10^-6
Using the quadratic formula [x = (-b ± sqrt(b^2 - 4ac)) / (2a)], we can calculate the values of x:
x = [-( -0.25) ± sqrt((-0.25)^2 - 4(1)(0.375 × 10^-6))] / (2(1))
x = [0.25 ± sqrt(0.0625 - 1.5 × 10^-6)] / 2
x = [0.25 ± sqrt(0.0614985)] / 2
x ≈ [0.25 ± 0.24798] / 2
Now we have two possible values for x:
1. x ≈ (0.25 + 0.24798) / 2
2. x ≈ (0.25 - 0.24798) / 2
Simplifying, we get:
1. x ≈ 0.24899 / 2 ≈ 0.1245
2. x ≈ 0.00202 / 2 ≈ 0.0010
Step 8: Determine the pH of the solution using the concentration of H3O+ obtained from the quadratic equation. Since we have found the concentration of H3O+ as approximately 0.1245 M, which is larger than 10^-7 M (the approximate concentration of H3O+ in pure water), the solution is acidic. Taking the negative logarithm (base 10) of the H3O+ concentration, we can find the pH:
pH ≈ -log(0.1245) ≈ 0.906
Therefore, the pH of the 0.25 M solution of the weak base is approximately 0.906.