Question

MOPS is often used as a buffer system in biochemistry. What is the pH of a buffer solution prepared by combining 0.200 moles of MOPS base and 0.120 moles of HCl in 2.00 L of water? (pKa = 7.20 for MOPS-H+)

The answer is 7.02

MOPS-Na + HCL → MOPS-H + NaCl

Answer 1

Hey how has your class been so far? I've been away from OS for sometime.

Answer 2

pH = pkA+Log[[A]/HA]

Answer 3

@Photon336 has given you the Henderson-Hasselbalch equation, which combines the pieces to make a useful buffer. A good buffer must have a weak acid (MOPS-H), and its conjugate (MOPS-), and both in approximately equal amounts if it's going to be effective.

The ratio of those two concentrations will change the pH of the buffer + or - relative to its pKa, depending on which component is there in greater concentrations.

Answer 4

@JFraser so when I plug in the numbers to the equation:

pH = pKa + log (base/acid)
pH = 7.20 + log (0.200/0.120)
pH = 7.42, which is too high

Answer 5

you've plugged in the \(moles\) of each, you need to find the \(concentration\) of each

Answer 6

well the concentration of HCl can be found by multiplying moles by L:
2.00 L * 0.120 moles = 0.24 M

but I don't know about the buffer

Answer 7

but you don't actually have the moles of MOPS-H, you're given the moles of HCl. You need to do the stoichiometry first to find out how much MOPS-H you form

Answer 8

Like this:

MOPS-Na + HCL → MOPS-H + NaCl
0.200 moles 0.120 moles x

Answer 9

basically, yes

Answer 10

how many moles of MOPS- are \(left over\), and how many moles of MOPS-H are \(formed\)?

Answer 11

I am not sure how you figure that out

Answer 12

this is a limiting reactant problem. If you start with 0.200mol of MOPS-, and 0.120mol of HCl, \(all\) of the HCl will be used, but only \(some\) of the MOPS- will be used, won't it?

Answer 13

moles of MOPS-H = 0.32
moles of MOPS left over is 0.32 - 0.200 = 0.12

Answer 14

no, the moles of MOPS-H formed is .0.120

Answer 15

How did you determine that?

Answer 16

because the products are formed, as the reactants are used up

Answer 17

MOPS-Na + HCL → MOPS-H + NaCl
0.200 moles 0.120 moles x
- 0.120 moles -0.120 moles + 0.120 moles
----------------------------------------------
0.08 moles 0 0.120 moles

Answer 18

is that what you mean?

Answer 19

MOPS left over is 0.08 moles
MOPS-H is 0.120 moles

Answer 20

exactly, yes

Answer 21

now find the concentrations of MOPS- (that's A-) and MOPS-H (that's HA) and solve

Answer 22

well to find concentration I know you want to multiply L by moles, but only 2.00 L of water was added

Answer 23

I am not sure where to go to from here.... I thought I would multiply both by 2.00 L, but that doesn't work

Answer 24

concentration is moles \(per\) liter, so you \(divide\) by the total volume of 2.00L

Answer 25

so,

0.08 moles / 2.00 L = 0.04 M
0.120 moles / 2.00 L = 0.06 M

pH = 7.20 + log (0.04 / 0.06)
pH = 7.02

Answer 26

it fits, GJ