A certain weak base has a Kb of 7.80 × 10-7. What concentration of this base will produce a pH of 10.17?
Well, isn't that basic, I mean, literally! To find the concentration of this weak base that will produce a pH of 10.17, we need to do some calculations.
First, let's convert the pH to pOH since they are inversely related. The pOH is given by pOH = 14 - pH. So, pOH = 14 - 10.17 = 3.83.
Now, let's find the concentration of OH- ions using the pOH value. We can do this by taking the antilog of the pOH. So, [OH-] = 10^(-pOH).
Substituting the pOH value, we have [OH-] = 10^(-3.83) = 4.6 × 10^(-4) (rounded to 2 decimal places).
Since this is a weak base, we can assume that the concentration of OH- ions is approximately equal to the concentration of the base itself. So, the concentration of the weak base would also be 4.6 × 10^(-4) M.
Voila! There you have it. The concentration of this weak base that will produce a pH of 10.17 is approximately 4.6 × 10^(-4) M.
To find the concentration of the weak base, we can use the pH of the solution and the Kb value.
1. Start by writing the dissociation reaction of the weak base in water:
weak base (B) + H2O ⇌ OH- + BH+
2. Since the base is weak, we can assume that the concentration of the base decreases by x, and the concentrations of OH- and BH+ increase by x.
3. Therefore, at equilibrium, the concentration of OH- is x, and the concentration of BH+ is x.
4. From the reaction, we can write the expression for Kb as:
Kb = [OH-][BH+] / [B]
5. Substitute the values into the equation:
7.80 × 10^-7 = (x)(x) / (initial concentration of base - x)
6. Since Kb is small, we can assume that the initial concentration of the base is equal to the concentration of OH-.
7. Simplify the equation:
7.80 × 10^-7 = x^2 / x
8. Rearrange to solve for x:
x = √(7.80 × 10^-7)
9. Calculate the value of x:
x ≈ 8.83 × 10^-4
10. Finally, the concentration of the weak base that will produce a pH of 10.17 is approximately 8.83 × 10^-4 M.
To determine the concentration of the weak base that will produce a pH of 10.17, we can use the concept of the dissociation of weak bases.
1. First, recall that the pH of a solution is related to the concentration of hydroxide ions (\(OH^-\)) present. In this case, we want to find the concentration of the base (\(B\)) that produces a pH of 10.17. Therefore, we need to find the concentration of \(OH^-\) ions at this pH.
2. The \(OH^-\) concentration can be calculated using the formula:
\[OH^- = 10^{-pOH}\]
where \(pOH\) is the negative logarithm of the hydroxide ion concentration.
3. Since we know the pH (\(pH = 10.17\)), we can find the pOH by subtracting the pH from 14 (pH + pOH = 14):
\[pOH = 14 - pH\]
\[pOH = 14 - 10.17 = 3.83\]
4. With the value of pOH, we can now calculate the concentration of \(OH^-\) ions using the formula for pOH:
\[OH^- = 10^{-pOH} = 10^{-3.83}\]
Solve this equation to find the concentration of \(OH^-\).
5. Now, since the weak base dissociates into one hydroxide ion (\(OH^-\)) for every one base molecule (\(B\)), we can conclude that the concentration of the base (\(B\)) is the same as the concentration of \(OH^-\). Therefore, the concentration of the weak base is the same as the concentration of \(OH^-\) ions calculated in step 4.
\[B = OH^- \quad \text{(concentration of weak base)}\]
6. Substitute the value of \(OH^-\) from step 4 into the equation above to find the concentration of the weak base.
pH = 10.17.
pOH = 14-10.17 = 3.83
pOH = -log(OH^-)
3.83 = etc.
OH^- = about 1.4E-4 but that's an estimate.
...........BH + H2O ==> BH2^+ + OH^-
I..........x............0........0
C.........-y............y........y
E.........x-y...........y.........y
Kb = 7.8E-7 = (BH2^+)(OH^-)/(BH)
You know OH^- = estimated 1.4E-4
You know BH2^+ = OH^- = y
Solve for x. You don't have two unknowns; only x is unknown.