a solution contains 0.216 moles of a base and 0.614 moles of the conjuagte acid in a total volume of 800 mL. The pH of the solution 9.65. what is the value for kb of the base?
the answer is 1.27E-4 how do you do it? do you need to convert from moles to molarity idek.
Use the Henderson-Hasselbalch equation and solve for pKa. Then remember that pKa + pKb = pKw = 14.
Solve for pKb.
Then pKb = -logKb and solve for Kb.
No, you're SUPPOSED to solve for M base and M acid but I never do because that's more work and the answer comes out the same. Note that M base is 0.216/0.800 and M acid is 0.614/0.800 and if you substitute the M in the HH equation you get the right answer. However, if you put that in the ratio of base to acid you have
(0.216/0.800)/(0.614/0.800) and you see the 0.800 value cancels. So you can do the work of dividing and calculating M but in the end just cancel the 0.800 and leave the mols.
i did not get the answer
You did something wrong. I got the answer. Post your work and I'll find the error
14-9.65=4.35 POH
log (0.27/.7675)= -.454
+ 4.35= 3.89
10^ (-3.89)= 1.28 E-4 M
A few comments. Your math is hard to follow. 4.35 certainly is not = 3.89
I would count 1.28E-4 as the same answer as 1.27E-4 but maybe your prof is tougher than that.
I think the reason your answer came out slightly larger (1.28 and not 1.27) is because you took the numbers out of the calculator and re-entered. If I leave the numbers in the calculator the answer comes out to be 1.27E-4. That's probably how the author of the problem did it with his/her calculator.
The 9.65 is not to be convert to pOH.
Here is how it is done.
9.65 = pKa + log(0.27/0.7675)
Do all of this in one step.
pKa = 10.104
pkb = 14-10.104 = 3.896
Kb = -log 3.896 = 1.2697E-4 which rounds to 1.27E-4. (I kept all of that in my calculator and just went from one step to the next and didn't round anything until the final step).
You did good work. The only two things wrong:
The conversion of 9.65 to pOH is not necessary and you can't use that number anywhere which caused the second small error of making 4.35 = 3.89. But you
had enough on the ball to use 3.89 and convert that to Kb and not to use 4.35 and convert that to Kb.
Hope this helps.
To find the value of Kb (the base dissociation constant) for the base, we need to use the information given and apply the concept of acid-base equilibria. Here's how we can solve it step by step:
1. First, we need to convert the given moles of the base and conjugate acid into molarity (M). To do this, we use the formula:
Molarity (M) = moles / volume (in liters)
Given that the total volume is 800 mL, we need to convert it to liters by dividing by 1000:
Volume = 800 mL / 1000 = 0.8 L
Now, calculate the molarity of the base and conjugate acid:
Molarity of base (M_base) = moles of base / volume
M_base = 0.216 moles / 0.8 L
Molarity of conjugate acid (M_acid) = moles of acid / volume
M_acid = 0.614 moles / 0.8 L
2. Next, we need to set up the equilibrium constant expression for the reaction of the base with water. For a base (B) reacting with water (H2O), the equilibrium equation is:
B + H2O ⇌ BH+ + OH-
The equilibrium constant expression for this reaction is:
Kb = ([BH+] * [OH-]) / [B]
Where [BH+] and [OH-] are the molar concentrations of the conjugate acid and hydroxide ion, respectively, and [B] is the molar concentration of the base.
3. We need to determine the molar concentration of hydroxide ions ([OH-]) from the pH of the solution. The pH of 9.65 tells us that the concentration of hydroxide ions is given by:
[OH-] = 10^(-pOH)
pOH = 14 - pH = 14 - 9.65 = 4.35
[OH-] = 10^(-4.35)
4. Now we can substitute the molar concentrations into the equilibrium constant expression and solve for Kb:
Kb = ([BH+] * [OH-]) / [B]
Since the conjugate acid concentration [BH+] is equal to [OH-] = 10^(-4.35), and the base concentration [B] is 0.216 M, we have:
Kb = (10^(-4.35) * 10^(-4.35)) / 0.216
Simplifying the expression gives:
Kb = 1.27E-4
Therefore, the value for Kb of the base is indeed 1.27E-4.