Question

Part A:

You are assigned to prepare a phosphate buffer with a pH = 7.70 using the 50mM Na2HPO4 and 50mM NaH2PO4 solutions in the lab. What is the ratio of conjugate base to 1 conjugate acid in this buffer? The pKa of H2PO4- is 6.82.

Part B:

Your group is assigned to prepare a phosphate buffer of pH = 7.70. What volume of 50. mM NaH2PO4 must be added to 25.00 mL of 50. mM Na2HPO4 to prepare this buffer? Calculate the answer to two decimal places and give the proper units in the answer (put a space between the numbers and the units).

@cuanchi
@Photon336
@JFraser

Answer 1

I solved part A correct, but I am only getting partial credit on Part B......

My work for Part A:

pH = pKa + log ([base] / [acid])
7.70 = 6.82 + log R
0.88 = log R
10^0.88 = R
7.59 : 1 = R

My work for part B:

The way I tried to solve it, and I only got partial credit was:

ratio = volume of the conjugate base / volume of conjugate acid
7.59 / 1 = 25.00 mL / X

7.59 X = 25.00 mL
X = 25.00 mL / 7.59
x = 3.30 mL

Answer 2

One sec, I'm just sorta refreshing my mind on how to do this so I ended up rederiving the Henderson-Hasselbach equation for part A there just to check that I can still do this and using your correct answer for it to check myself. So far so good haha.

Answer 3

I am trying to find a better way to explain this, but in the mean time, maybe you can figure out how to do it better than me.

You're not including the volume change that comes from the addition of acid, it increases, it should be more than 25.00mL slightly.

Answer 4

idk, maybe I'm wrong. I'll try to find out since I should know this.

Answer 5

Ok so from part A you used the fact from the Henderson-Hasselbach equation:

\[10^{(pH-pK_A)} = \frac{[HA]_f}{[A^-]_f}\]

that's true at the final concentration, so I added little f subscripts to make it stand out. The final volume is the same for both, since they're in the same beaker at this point, so we can multiply by a fancy form of 1:


\[10^{(pH-pK_A)} = \frac{[HA]_fV_f}{[A^-]_f V_f}\]

We don't know what the final volume is, but we know that the number of moles of base stays the same: \([A^-]_fV_f=[A^-]_iV_{A^-, i}\) We know the starting concentration and starting volume of base, which is great, so it's stuff we know. We can go ahead and do some algebra after plugging that in to get:

\[[A^-]_iV_{A^-, i}10^{(pH-pK_A)} = [HA]_fV_f\]

Now we do something similar with the acid, \([HA]_fV_f=[HA]_iV_{HA, i}\)
Aha, now we have the initial acid volume (what we wanted all along) and we know the initial acid concentration too! So we now can plug this in and we have everything we need to solve.

\[\frac{[A^-]_iV_{A^-, i}}{[HA]_i}10^{(pH-pK_A)} = V_{HA, i}\]
\[\frac{(50mM)(25.00mL)}{50 mM} 10^{(7.70-6.82)} = V_{HA, i}\]
\[189.6mL = V_{HA,i}\]
This seems like substantially more than what you put. Let's see if this actually works out by plugging it back into the HH equation.

\[pH = pK_A + \log \frac{[HA]}{[A^-]}\]
Ok moment of truth, plugging in initial volumes and concentrations:
\[7.70 = 6.82 + \log \frac{(189.6 mL)(50 mM)}{(25.00mL)(50mL)}\]
It actually works out, I'm shocked haha.

Answer 6

OK so "plug and chug" method would be to start here:

\[pH = pK_A +\log_{10} \frac{[HA]_f}{[A^-]_f}\]
and then plug in initial volumes and concentrations, knowing that their final volumes are equal.


\[pH = pK_A +\log_{10} \frac{[HA]_iV_{HA,i}}{[A^-]_iV_{A^-,i}}\]

in this case your only unknown was the initial acid volume, \(V_{HA,i}\). Sorry for the confusing looking subscripts.

Answer 7

If it helps, the initial separated volumes of acid and base are the final volume.
\[V_f= V_{HA, i} + V_{A^-,i}\]

Answer 8

@Kainui

For Henderson-Hasselbach equation, The base needs to be the numerator and the acid should be the denominator. The question is asking for the volume of NaH2PO4, which is the acid must be added to the 25.00 mL of the base which is Na2HPO4..... so the whole equation needs to be flipped

Answer 9

When you are checking the answer it should be:
pH = pKa + log [A-] / [HA], in your equation you have the acid over the base.... which is wrong

so when the volumes are plugged in correctly.... I don't get 7.70 as the pH.... I get 5.94.... so I am unsure that that is the correct volume.... ?

Answer 10

What I get is:

\[pH = pKa + \log _{10} \frac{ [base] V _{base} }{ [acid] V _{acid} }\]

so I would solve for V acid:

\[7.70 = 6.82 + \log \frac{ [50 nM] * 25.00 mL }{ [50 nM] * 3.26 mL }\]