0
$\begingroup$

It's been a while since I've been in a chemistry class and I'm having a hard time figuring this out. I need to reduce the pH of a solution from 5 to 4. There are 3 acids I can use to do this: 20% v/v sulfuric acid, concentrated hydrochloric acid (assume 36% w/w), or glacial acetic acid. I calculated the molarity of each in the table below using density and molecular weight. For the calculations I'll assume the solution I need to acidify is 1 liter, but I'll actually be scaling this up to a large tank.

Value 20% v/v H2SO4 Concetrated HCl (36% w/w) Glacial Acetic Acid
mol/L 3.73 11.75 17.42
pKa1 -10 -7 4.76
pKa2 1.99

The starting solution has $[H^+]_1 = 10^{-pH_1} = 10^{-5} \frac{mol}{L}$, and I need to increase it to $[H^+]_2 = 10^{-pH_2} = 10^{-4} \frac{mol}{L}$. For a 1 liter solution, that's $(10^{-4} \frac{mol}{L} - 10^{-5}\frac{mol}{L})*1L = 9.0x10^{-5} mol\,H^+$.

I think I got the HCl alright because it's a strong acid, so 1 mol HCl $\rightarrow$ 1 mol H+. Therefore $9.0x10^{-5} mol\;H^+*\frac{1L}{11.75mol}=7.66x10^{-6}L$ HCl solution needed.


For the H2SO4, if I ignore the weak acid HSO4-, then I can do the same calculation as for HCl and get
$9.0x10^{-5} mol\;H^+*\frac{1L}{3.73mol\;H_2SO_4}=2.41x10^{-5}L$.
But I'm not sure if I can ignore the second proton since the pKa2 is still pretty low at 1.99. Calculating [H+] from HSO4- than I get $K_{a2}= \frac{[HSO_4^{2-}][H^+]}{[HSO_4^-]} \rightarrow 10^{-1.99}=\frac{x^2}{3.73M} \rightarrow x=0.195M=[H^+]$. Does that mean that 1 L of 3.73 molar H2SO4 will contribute 3.73 + 0.195 = 3.925 mol H+? In that case I'd need
$9.0x10^{-5} mol\;H^+*\frac{1L}{3.925mol}=2.29x10^{-5}L$ of the H2SO4 solution. That's 5% difference, which will be a lot when I scale this up.


For the glacial acetic acid, I can do the same calculation I did for the HSO4-.
$K_a= \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \rightarrow 10^{-4.76}=\frac{x^2}{17.42M} \rightarrow x=0.0174M=[H^+]$.
$9.0x10^{-5} mol\;H^+*\frac{1L}{0.0174mol}=0.0517L$ of the acetic acid solution.


So to summarize, to decrease the pH of a 1 liter solution from 5 to 4, I need to add 9.0x10^-5 mol [H+], or
7.66 microliters of 11.75M HCl,
24.1 microliters of 3,73M H2SO4, or
51.7 milliliters of 17.42M CH3COOH (glacial acetic acid).

Is this correct, or am I missing something? Now the kicker - how does this change if the solution I'm trying to acidify is buffered? What do I need to know about the buffer in order to calculate the pH change with these 3 acids?

$\endgroup$
4
  • $\begingroup$ All your calculations only make sense if you know the nature of the original solution. You have to know the formula or the pKa of the initial acid. If the initial solution with pH 5 contains a strong acid, the calculation is rather different from the calculation made with a solution of a weak acid. $\endgroup$
    – Maurice
    Mar 11, 2022 at 17:24
  • $\begingroup$ Are my calculations correct in the case of a strong acid solution, ie not buffered? I think I can determine the pKa of the pH 5 solution experimentally via titration, but how would the calculations change if it is a weak acid? $\endgroup$
    – Brian
    Mar 11, 2022 at 18:31
  • $\begingroup$ What is the background of your question? If you are away of education, you are unlikely to do such tasks, that seem to be made up for sake of education. $\endgroup$
    – Poutnik
    Mar 13, 2022 at 7:28
  • $\begingroup$ I work for a small manufacturer that mostly relies on pre-mixed proprietary chemical mixes, but does some R&D here and there. Some chemical suppliers are willing to help with these sorts of inquiries, but others can't be bothered. $\endgroup$
    – Brian
    Mar 17, 2022 at 17:14

1 Answer 1

1
$\begingroup$

If the solution is unbuffered, you could ignore that it has a pH of 5 and start with pure water instead for a rough estimate.

The calculation for $\ce{HCl}$ makes sense. The calculation for $\ce{H2SO4}$ does not; you know that the final pH will be 4. At pH = 4, sulfuric acid will have lost 2 protons (you should have used 0.0001 for the $\ce{H+}$ concentration rather than "x"). So you need half the amount of sulfuric acid compared to hydrochloric acid.

In practical terms, you would hardly be able to add exactly 7.66 $\mu$L HCl to your solution. Typically, you would get a rough estimate of how much you need to add and then add an appropriately concentrated solution dropwise until you reach the desired pH.

If the solution is buffered, you need to know the concentration of the buffer substances along with their pKa values. Or you would need to know the buffer capacity between pH 5 and 4. In this scenario, you might take a small sample and adjust the pH, giving you a sense how much you need to add to the entire solution without zipping past the desired pH.

$\endgroup$
1
  • $\begingroup$ Thanks for your explanation. Dealing in μL is a bit ridiculous, but this is just a feasibility calculation for a large chemical tank which is 100s of gallons. I'll probably end up trying to adjust a small sample like you suggested, but I also want to understand what's going on chemically, not just operationally. $\endgroup$
    – Brian
    Mar 17, 2022 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.