The pH of a 0.065 M solution of a weak base is 11.70. What is the value of Kb for this base?
You're cheating on an assignment due for the class Chemistry 1012, your chemistry professor will be notified.
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they cant prove its you, plenty of people get the same problem, and finding help is not cheating....
-log(PH)=H
-log(POH)=OH
[OH][H]/[HA]=Ka
Kw=1*10^-14
Kw/Ka=Kb
those are the equations you need, just solve it... I don't see the cheating in help and I'M A TEACHER.
That's why you just shouldn't use your real name
To find the value of Kb for the weak base, we need to use the relationship between pH, pOH, and pKw (the ion product constant for water). Here are the steps to find the value of Kb:
Step 1: Calculate pOH
Since we know the pH of the solution, we can calculate the pOH using the formula:
pOH = 14 - pH
In this case, the pH is 11.70, so:
pOH = 14 - 11.70 = 2.30
Step 2: Calculate [OH-]
To find the concentration of hydroxide ions ([OH-]), we need to convert pOH to OH- concentration using the formula:
[OH-] = 10^(-pOH)
In this case:
[OH-] = 10^(-2.30)
Step 3: Calculate [OH-]/[B]
In a 0.065 M solution of the weak base (let's call it B), the concentration of the base is equal to [B].
[B] = 0.065 M
Now, divide the concentration of hydroxide ions ([OH-]) by the concentration of the weak base ([B]):
[OH-]/[B] = [OH-]/[B] = [OH-] / 0.065
Step 4: Calculate Kb
Kb is the equilibrium constant for the reaction of the weak base with water, and it can be calculated using the following formula:
Kb = [OH-][OH-]/[B]
Now, substitute the known values to calculate Kb:
Kb = ([OH-] / 0.065) * ([OH-] / 0.065)
Kb = ([OH-] ^ 2) / (0.065 ^ 2)
Finally, plug in the value of [OH-]:
Kb = (10^(-2.30) ^ 2) / (0.065 ^ 2)
Once you solve this expression, you will find the value of Kb for the weak base.