Question

A beaker with 115mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 4.00mL of a 0.490M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Answer 1

this is a fairly long problem, are you there still?

Answer 2

I'm here now. This problem is still confusing me.

Answer 3

any ideas on starting this one?
are you familiar with the Henderson Hasselbalch equation? @llaurenemilyy

Answer 4

Yeah, I am. I kinda know the basics, it's just a different problem than I've been working on.

Answer 5

and we will have two sets of conditions right?

Answer 6

Yeah, I think so.

Answer 7

right there is some algebra to this
we can plug in what we know right?
the first condition is that base + acid = .1M right?

Answer 8

so base = .1 - acid
by rearranging
then we know that pka is given right?

Answer 9

Yes. That's right.

Answer 10

we can then solve for the concentration of base and acid right?

Answer 11

5.00 = 4.760 + log ((.1-acid)/acid)

5.00 = 4.760 + log ((.1-acid)/acid))
okay, so we can solve for acid right?

Answer 12

Yeah, you can use x=acid. And solve for x.

Answer 13

subtracting 4.76 from each side
and then we can undo the log by taking 10^ for each side right?

Answer 14

Correct.

Answer 15

and yes you're right!

Answer 16

So 10^-0.24
x=0.567

Answer 17

now any ideas on solving for the concentration of base?

Answer 18

0.576

Answer 19

That's what I meant. Haha

Answer 20

hmm how did you get that though?
remember we have ((.1-acid)/acid) on the right side

Answer 21

Should it be 10^0.24 or 10^-0.24?

Answer 22

10^.24 right?
since 5.00-4.74 is .24

Answer 23

Okay, that makes sense.

Answer 24

okay, are you getting the same value for the acid as I did?
and now we can solve for the base based on .1 = base + acid

Answer 25

Yeah, I did. Base is 0.0635M

Answer 26

Exactly, nice work! :]

Answer 27

okay, now we know the initial conditions right?
the concentration of the acid and conjugate base in the buffer

Answer 28

so we have 115mL of those concentrations
can you find the moles of acid and base?

Answer 29

that way we can see what affect the 0.490M HCl solution will do to those moles

Answer 30

Yes, turn mL to L and then multiply M and L.

Answer 31

exactly

Answer 32

.115* .0365 = moles acid .0041975 moles acid
.115* . 0635= moles base .0073 moles base

Answer 33

0.0042 mol acid & 0.0073 mol conjugate base

Answer 34

Okay, that's what I got!

Answer 35

Great!

Answer 36

okay, and then we need to find moles of HCl added right?

Answer 37

Yeah, so do the same procedure.

Answer 38

.004* .490 moles HCL .00196 moles HCL

Answer 39

okay, so we have these values
have you done limiting reactant type problems before?
we have the reaction base + HCl --> acid right?
each mole of HCl will convert 1 mole of the conjugate base back to the acetic acid

Answer 40

Yeah, that's what I got! And yes I have done problems like that before.

Answer 41

Awesome!

Answer 42

so lets find out how much of the base is converted to acid by the HCl

Answer 43

Do you add the mols of base and moles of HCl, then subtract mols of acid? I'm not sure if I know how to do that.

Answer 44

nopewe need to subtract like I have on the doc
the HCl uses up the base, until the acid runs out and then the acid will be produced by the HCl.

Answer 45

Oh Makes sense!

Answer 46

so what would the final values be? - referring to the doc!

Answer 47

and then we can plug these directly back into a new henderson-hasselbalch equation

Answer 48

Final base= 0.00534 and final acid= 0.00224

Answer 49

Oh wait, for the acid I subtracted it. It needs to be added.

Answer 50

oh
the acid gets added right?

Answer 51

yes, glad you picked up your mistake!

Answer 52

Yeah, the acid=0.00616, right?

Answer 53

Yes!.... the acid gets added
.0041975 + .00196
because the HCl converts the base to acid

How might that new H-H equation look? we don't need to find molarities
since the volumes of both are the same and they get divided.Final base= 0.00534 and final acid= 0.00224 so we can plug the moles straight in

Answer 54

and now we plug into H-H equation right?
we know pka and this ratio of base/acid
so we can solve for pH

Answer 55

pH=4.70

Answer 56

yes, does that make sense? :]

Answer 57

It does! Thank you so much! You were such a great help!

Answer 58

No worries, glad I helped! Good luck! :-]

Answer 59

:]

Answer 60

Thank you again!

Answer 61

yw :]