Describe using complete calculations how you would prepare 20.0 mL of a buffer with a pH of 9.00 using 0.12 M NH4Cl (aq) and 0.15 M NH3 (aq) solutions. The total concentration of NH3 plus NH4+ should be 0.10 M. Kb(NH3) = 1.8 x 10^-5.
No idea how to approach this question. Thanks I’m advance!
eqn1 is pH = pKa NH3 + log (base)/(acid)
I think pKa for NH3 is 9.26 but you should make sure.
9 = 9.26 + log b/a
solve for b/a
eqn2 is a + b = 0.1 M
Solve these two equations simultaneously for acid and base
That will tell you what (acid) and (base) must be in the final solution. Is this enough to get you started.? Post your work if you get stuck and/or need more help.
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Here's my work:
Ka = Kw / Kb
= 1x10^-14 / 1.8x10^-5
= 5.55x10^-10
pKa = -log(5.55x10^-10) = 9.26
pH = pKa + log(b/a)
9 = 9.26 + log(b/a)
log(b/a) = -0.26
b/a = 0.55
b = 0.55a
a + b = 0.10 M
a + 0.55a = 0.10
a = 0.1 / 1.55
a = 0.0645 M
b = 0.1 - 0.0645 = 0.0355 M
Volume of NH4Cl needed:
0.0645 x 0.2 / 0.12 = 0.1075 L = 107.5 mL
Volume of NH3 needed:
0.0355 x 0.2 / 0.15 = 0.0473 L = 47.3 mL
And 200 - 107.5 - 47.3 = 45.2 mL of H2O
Is this correct? Thanks for all the help you've given!
Unrelated question: do you folks get notifications whenever a question you've answered gets new replies?
Your solution looks OK to me.
On the related question, technically we don't get notified of anything. The post you made is there for everyone to see as is my reply. If you reply to that and I go back and look then your next replies, as well as any other posts is there for me and anyone else to see. I think most of the tutors here go back a bit later and look to see if a follow up is needed. I was away from my desk until about 10:30 my time (CT) so there is a lag sometimes in follow up times. We don't have set times to help; it's just when we feel like it since all of us are volunteers and we do this for the fun of it.You did a great job. Thanks for using Jiskha
Sorry, correction: I meant 200.0 mL, not 20.0.
To prepare a buffer solution, we need to mix a weak acid and its conjugate base or a weak base and its conjugate acid. In this case, we are dealing with a weak base (NH3) and its conjugate acid (NH4+). The Henderson-Hasselbalch equation can be used to calculate the ratio of the concentrations of the weak acid and its conjugate base in order to achieve the desired pH.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
where pH is the desired pH, pKa is the logarithmic acid dissociation constant, [A-] is the concentration of the conjugate base (NH3), and [HA] is the concentration of the weak acid (NH4+).
First, we need to determine the pKa of NH4+. Since NH3 and NH4+ are conjugate acid-base pairs, we can use the pKa and pKb relationship:
pKa + pKb = 14
Given that Kb(NH3) = 1.8 x 10^-5, we can calculate the pKa:
pKb = -log(Kb(NH3))
pKa = 14 - pKb
pKa = 14 - (-log(1.8 x 10^-5))
Now that we have the pKa, we can use the Henderson-Hasselbalch equation to calculate the ratio of [A-]/[HA]:
9.00 = pKa + log([A-]/[HA])
Solving this equation for log([A-]/[HA]) gives:
log([A-]/[HA]) = 9.00 - pKa
Now, we can calculate the concentration of NH3 and NH4+ needed to achieve the desired pH and total concentration.
First, determine the concentration of NH4+ using the total concentration and the concentration of NH3:
0.10 M = [NH3] + [NH4+]
Given that [NH3] + [NH4+] = 0.10 M, we can substitute [NH3] in terms of [NH4+] using the Henderson-Hasselbalch equation:
[NH3] = (10^log([A-]/[HA])) * [NH4+]
Since [NH3] + [NH4+] = 0.10 M, we can substitute [NH3] and solve for [NH4+]:
0.10 M = (10^log([A-]/[HA])) * [NH4+] + [NH4+]
0.10 M = [NH4+](1 + 10^log([A-]/[HA]))
Now, we substitute the determined values into the equation:
0.10 M = [NH4+](1 + 10^(9.00 - pKa))
Solving this equation for [NH4+] gives us the concentration of NH4+ needed to prepare the buffer solution.
Once we have the concentration of NH4+, we can determine the volume of the NH4Cl solution needed to achieve the desired concentration using the following equation:
M1V1 = M2V2
Where,
M1 = concentration of NH4Cl (0.12 M)
V1 = volume of NH4Cl solution (unknown)
M2 = concentration of NH4Cl needed (determined from the equation above)
V2 = final volume of the buffer solution (20.0 mL)
Substituting the values into the equation, we can solve for V1:
(0.12 M)(V1) = (NH4+ concentration) (final volume)
V1 = (NH4+ concentration) (final volume) / 0.12 M
Now, you can substitute the calculated values to determine the required volume.
Note: It is important to double-check your calculations and perform the necessary unit conversions to ensure accurate results.