A certain weak base has a Kb of 8.10 × 10-7. What concentration of this base will produce a pH of 10.00?
pH = 10; pOH = 4; (OH^-) = 1E-4
.............BOH ==> B^+ + OH^-
initial.......x......0......0
equil.......x-1E-4...1E-4..1E-4
Kb = (B^+)(OH^-)/(BOH)
To determine the concentration of the weak base that will produce a pH of 10.00, we need to use the relationship between pH and the concentration of hydroxide ions (\(OH^-\)).
The pH scale is calculated using the formula: \(pOH = -log[OH^-]\), where [OH^-] is the concentration of hydroxide ions in moles per liter (M).
Since we are given a pH of 10.00, we can calculate the pOH value:
\(pOH = 14 - pH = 14 - 10.00 = 4.00\)
The pOH is related to the concentration of hydroxide ions by the formula: \(pOH = -log[OH^-]\). Rearranging the equation, we get:
\([OH^-] = 10^{-pOH}\)
Substituting the value of pOH, we have:
\([OH^-] = 10^{-4.00} = 0.0001\, M\)
Since the weak base (\(B\)) reacts with water (\(H_2O\)) to produce hydroxide ions (\(OH^-\)) according to the equation:
\(B + H_2O \rightarrow BH^+ + OH^-\)
We can assume that the concentration of the weak base (\(B\)) is equal to the concentration of hydroxide ions (\(OH^-\)).
Therefore, the concentration of the weak base that will produce a pH of 10.00 is 0.0001 M.
To determine the concentration of the weak base that will produce a pH of 10.00, we need to use the equation for the dissociation of a weak base in water:
B + H2O ⇌ BH+ + OH-
The equilibrium constant for this reaction is defined as the base dissociation constant, Kb:
Kb = [BH+][OH-] / [B]
Given that the Kb of the weak base is 8.10 × 10^(-7), we can set up the equation as follows:
8.10 × 10^(-7) = ([BH+][OH-]) / [B]
Since the pH is 10.00, we know that the concentration of hydroxide ions ([OH-]) is 10^(-4) M (because pH = -log[OH-], so [OH-] = 10^(-pH)).
Now we can substitute the known values into the equation:
8.10 × 10^(-7) = (x × 10^(-4)) / [B]
where x represents the concentration of the weak base ([BH+]).
To find the concentration of the weak base, [B], we need to rearrange the equation:
[B] = (x × 10^(-4)) / 8.10 × 10^(-7)
[B] = x × (10^(-4) / 8.10 × 10^(-7))
Simplifying further:
[B] = x × 10^(3)
At equilibrium, the concentration of the weak base is equal to the concentration of the conjugate acid ([BH+]). Therefore, we can replace [BH+] with x:
[B] = x × 10^(3)
Since the concentration of the weak base is usually much larger than the concentration of hydroxide ions, we can assume that the change in concentration of the weak base is negligible compared to the original concentration, so [B] ≈ [BH+]. Thus, we can also replace [B] with x:
x = x × 10^(3)
To solve for x, we divide both sides of the equation by x:
1 = 10^(3)
Here, we observe that there is no value of x that satisfies the equation. This means that the concentration of the weak base needed to produce a pH of 10.00 is not achievable with a Kb of 8.10 × 10^(-7).