A certain weak base has a Kb of 9.00x10^-7. What concentration of this base will produce a pH of 10.30?
pH = 10.3 = -log(H^+)
(H^+) = 5.01E-11 and (OH^-) = 2E-4
.........B + HOH ==> BH^+ + OH^-
I........x............0.......0
C........-y...........y.......y
E........x-y..........y........y
y = 2E-4
Ka = 9.00E-7 = (BH^+)(OH^-)/(B)
(BH^+) = (OH^-) = 2E-4
(B) = x-2E-4. Solve for x-y
To find the concentration of the weak base that will produce a pH of 10.30, we need to use the relationship between pH, pOH, and pKw.
First, let's calculate the pOH using the formula:
pOH = 14 - pH
Given that the pH is 10.30, we have:
pOH = 14 - 10.30 = 3.70
The pKw of water is defined as:
pKw = pH + pOH
Since we know the pOH and pH, we can substitute them into the equation:
pKw = 10.30 + 3.70 = 14
Now, we can use the pKw value to find the concentration of the weak base.
pKw = pKa + pKb
Since we are given the Kb value as 9.00x10^-7, we can find the pKb value:
pKb = -log(Kb) = -log(9.00x10^-7)
Calculating this using a calculator, we find pKb ≈ 6.05
Now that we know the pKb value, we can use it to find the concentration of the weak base using the equation:
Kb = [OH-][HB]/[B-]
Since the weak base dissociates into its conjugate acid (HB) and hydroxide ions (OH-), we can write the expression as:
Kb = [OH-]^2/[B-]
Rearranging the equation, we get:
[OH-]^2 = Kb * [B-]
Now, substitute the given pOH value into the equation to find [OH-]:
[OH-] = 10^(-pOH)
[OH-] = 10^(-3.70)
Calculating this using a calculator, we find [OH-] ≈ 0.0199
We have [OH-], and we can find [B-] from the equation:
[B-] = [OH-] * Kb
Substituting the values:
[B-] = 0.0199 * 9.00x10^-7
Calculating this, we find [B-] ≈ 1.79x10^-8
Thus, the concentration of the weak base that will produce a pH of 10.30 is approximately 1.79x10^-8.