A certain weak base has a Kb of 8.40 × 10-7. What concentration of this base will produce a pH of 10.34?

pH = 10.34

pOH = 14-10.34 = 3.66
Then (OH^-) = about 2E-4 but that's only an estimate.
........BOH ==> B+ + OH^-
I........x.......0.....0
C.......-y.......y.....y
E.......x-y .....y.....y

Kb = 8.4E-7 = (B^+)(OH^-)/(HB)
You know y and Kb, substitute into Kb expression and solve for x.

To find the concentration of the weak base that will produce a pH of 10.34, we can use the equation for the dissociation of the weak base, which follows the general form:

Base + H2O ⇌ OH- + Acid

The equilibrium expression for this dissociation is:

Kb = [OH-][Acid] / [Base]

Since Kb is given as 8.40 × 10^(-7), we can assume that the concentration of the OH- ion is equivalent to the concentration of the weak base [Base]. Thus, we have:

Kb = [OH-]^(2) / [Base]

Rearranging the equation to solve for [Base], we get:

[Base] = [OH-]^(2) / Kb

Now, to determine the concentration of OH-, we need to calculate the pOH using the pH value given. The relationship between pH and pOH is:

pH + pOH = 14

So, pOH = 14 - pH = 14 - 10.34 = 3.66

Finally, we can find the concentration of OH- using the pOH value:

[OH-] = 10^(-pOH)

[OH-] = 10^(-3.66)

[OH-] = 2.58 × 10^(-4) M

Substituting this value into the previous equation, we can calculate the concentration of the weak base ([Base]):

[Base] = [OH-]^(2) / Kb

[Base] = (2.58 × 10^(-4))^2 / (8.40 × 10^(-7))

[Base] = 1.032 × 10^(-4) M

Therefore, the concentration of the weak base that will produce a pH of 10.34 is approximately 1.032 × 10^(-4) M.

To determine the concentration of the weak base that will produce a pH of 10.34, we need to use the equation for the relationship between pH and pOH of a solution:

pH + pOH = 14

First, calculate the pOH using the given pH value:

pOH = 14 - pH
pOH = 14 - 10.34
pOH = 3.66

We know that pOH is related to the concentration of the hydroxide ions (OH-) in a solution. The concentration of OH- ions can be determined using the expression for the equilibrium constant for the base:

Kb = [OH-][HB] / [B-]

Where:
Kb = base dissociation constant
[OH-] = concentration of hydroxide ions
[HB] = concentration of the conjugate acid (if present)
[B-] = concentration of the base

Since we know the value of Kb (8.40 × 10^-7) and we are interested in finding the concentration of the base, we can assume that the concentration of the conjugate acid ([HB]) is negligible compared to [B-]. Thus, we can simplify the equation:

Kb ≈ [OH-][B-]

Next, substitute the value of pOH and the expression for Kb into the equation and solve for [OH-]:

Kb = [OH-][B-]
8.40 × 10^-7 = [OH-][B-]

Now, convert pOH to OH- concentration:

pOH = -log[OH-]
[OH-] = 10^(-pOH)
[OH-] = 10^(-3.66)

Finally, substitute the value of [OH-] into the equation to find the concentration of the base ([B-]):

8.40 × 10^-7 = 10^(-3.66)[B-]

To isolate [B-], divide both sides of the equation by 10^(-3.66):

[B-] = (8.40 × 10^-7) / 10^(-3.66)

[B-] ≈ 8.185 × 10^(-4)

Therefore, a concentration of approximately 8.185 × 10^-4 M of the weak base will produce a pH of 10.34.

I don't understand what Y and X are supposed to be