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!
4 answers
Sorry, correction: I meant 200.0 mL, not 20.0.
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.
.
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.
.
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?
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
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