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?