# Wondering how I should have graded this question (freshman lab, about buffer capacity)

I TA the intro freshman lab course. In a lab about buffers and buffer capacity, the students were instructed to make 50 mL of a 0.1 total molar pH 6.70 buffer by dissolving solid $KH_2PO_4$ and $K_2HPO_4$. The students then had to add a drop or two of NaOH and HCl to the buffer and see if/how much the pH changes. Next, they are supposed to dilute the buffer to double volume with DI water, and repeat. Then respond to the following questions:

1: How did diluting the buffer effect the change in pH when a few drops of NaOH or HCl were added?

2: Did adding water to the buffer change its pH? Explain why or why not.

For #1, obviously, diluting the buffer lowers its buffer capacity, so the pH change in the diluted buffer will be more drastic. I was a bit unsure about how to grade #2, however. Some of the answers I got were:

"The pH raised slightly but was small enough that it was probably random fluctuations in the pH meter"

"We didn't measure a pH change"

"The pH went up as DI water has a pH of 7 which is above the buffer, so even though the buffer resists change, it'll change a little"

If you look at the Henderson-Hasslebalch equation:

$pH = pKa + \log(\frac{[A^-]}{[HA]})$

Since both pH of the buffer was 6.7, and the pKa of the system is also 6.7, $[A^-]$ and $[HA]$ must be equal, and since the total concentration was 0.1M, then $[A^-]$ and $[HA]$ are 0.05 M each. So if we dilute it to double volume with water, the concentrations become 0.02 M each; the ratio remains unchanged and the pH would still be 6.7.

In reality, when you dilute the solution, the Ka and therefore pKa of the $KH_2PO_4 \leftharpoonup\rightharpoondown K_2HPO_4$ equilibrium will change slightly, so the pH would be expected to change. For clarification, see this question: Does dilution of a buffer affect pH?

The issue is that this is an early lab in a freshman course and they certainly aren't expected to know about activity vs concentration or anything that complex yet. There is no answer key and I am not TA for the lectures, so I am not expected to go to them, so I don't know what they were told there, or in their textbook. My supervisor seemed to give a non-answer when I asked about this. So I was a bit unsure what kind of thing I should be expecting the students to write. The lab will be the same next semester and I want to know what sort of answers to accept.

• Why does dilution affect capacity? The amount of conjugate base available to neutralize acid is the same. In strong dilutions, the capacity may be affected once water autoionization becomes a major factor in the equilibrium. Also, $K_{\mathrm{a}}$ is a property of an acid and not subject to change. – Zhe Jun 21 '17 at 18:13
• Say you have a a 1 molar total (so 0.5 molar for HA and A) buffer, pKa = 6.7, pH = 6.7. Add 0.1 moles of HCl, so 0.1 moles of A- are converted into HA. $pH = 6.7 + \log(\frac{[.5-.1]}{[.5+.1]}) = 6.524$ Technically in the brackets should be concentration, not moles, but since they are in the same solution the volume is the same so that cancels out. Now say the buffer was diluted to double volume, so now the concentrations of A and HA are 0.25, and you repeat: $pH = 6.7 + \log(\frac{[.25-.1]}{[.25+.1]}) = 6.332$ So the diluted buffer changes more, as it has less capacity. – iammax Jun 21 '17 at 18:35
• That's not correct because 0.1 moles of $\ce{HCl}$ would need to be converted to a concentration if you want to add or substract from concentration in solution... And technically, you can just leave the values of $\ce{HA}$ and $\ce{A-}$ at moles because the total volume is the same and those will cancel in the fraction. – Zhe Jun 21 '17 at 19:11
• Would you say that when you take 1 mL of a 10 mL buffer, the 1 mL has a lower buffer capacity? – Karsten Theis Jan 4 at 16:00