pH calculations, 4 (SB-62) One liter of two solutions A and B with the same starting pH of 4.8 shall be given. A is a hydrochloric acid solution, B is an acidic acid solution. Now one liter of water is added to each solution. Calculate the new pH of solution A and B In the case of a solution that has a greater change in pH, Note: HCl is a strong acid (pKs < 0), acetic acid a weak acid (pKs > 0)

  Exercise: SB-62    

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One liter of two solutions A and B with the same starting pH of 4.8 shall be given. A is a hydrochloric acid solution, B is an acidic acid solution.
Now one liter of water is added to each solution.

1. Calculate the new pH of solution A and B
2. In the case of a solution that has a greater change in pH,

Note: HCl is a strong acid (pKs < 0), acetic acid a weak acid (pKs > 0)

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Solution:

Based on the pH value, HCl for the strong acid and acidic acid for the weak acid can now be the concentration is calculated:

1. strong acid: HCl
   pH = -log(c(H₃O⁺)), c(H₃O⁺) = 10^-pH = 0.000015 mol/l
   C = n/V or n = c · V
   n = 0.000015 · 1.0 l = 0.000015 mol
2. Weak acid: acetic acid
   pH = 1/2(pKs-log(HA)), c(HA) = …
   n = c·V =
3. Now 1 litre of pure water is added to each solution, and with c=n/V we get the following:
4. HCl: pH = -log(c(H₃O⁺)) = -log(n/V)
   pH = [translation unavailable]

… not finished yet !

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