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pH calculations, 4 (SB-62)

One liter of two solutions A and B with the same starting pH of 4.8. 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,

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

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

1. High acidity: 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 every solution becomes 1 L of pure water is admitted, with c=n/V resulting the following:
4. HCl: pH = -log(c(H₃O⁺)) = -log(n/V)
   pH = log

… not finished yet !