It holds at the same time, equivalence point: n(Base) = n(Säure)
with c=n/V we get:
c(Base)·V(Base) = c(Säure)·V(Säure)
c(Säure) = c(Base)·V(Base)/V(Säure)
c(Säure) = 0.1M · 30ml / 20 ml = 0.15 M = 0.15 mol/l

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 !

Titration of an acid (20 ml, unknown concentration) with a base (0.1 M) is required up to the equivalence point of 30 ml of the base. Calculate the concentrations of the acid.

1. c=n/V → n=c·V
       · 80 ml 0.05 M HCl → n(HCl)=0.004 mol
       · 100 ml 0.01 NaOH → n(NaOH)=0.001 mol
       · Remaining staying at the neutralization reaction 0.003 mol HCl
       · Note: total volume is 80ml + 100 ml
       · pH=-log(0.003/0.18) = 1.78
2. · 60 ml 0.015 M KOH → n(KOH)=0.0009 mol
       · 30 ml 0.2 M HCl → n(HCl)=0.006mol
       · Note that KOH behaves similarly to NaOH, so it's also a base.
       · Remaining staying at the neutralization reaction 0.0051 mol HCl
       · Please note: total volume 60ml + 30 ml
       · pH=-log(0.0051/0.09) = 1.25
3. · Start pH: pH=-log(0.2) = 0.7
       · Tenfold dilution I mean,, that to the 10 ml still additional 90 ml this is the. Water is given
       · 10 ml 0.2 M HCl → n(HCl)=0.002 mol or 0.002 mol
       · n(HCl)=0.002 mol → 0.002 mol H⁺
       · Total volume: 10ml+90ml=100 ml
       · pH=-log(0.002/0.1) = 1.7
       · Increase by one unit of pH makes sense at ten times dilution
      
   

   ---

       · 1 g Ca(OH)₂ → n(Ca(OH)₂)=1/74.1=0.0135 mol
       · n(Ca(OH)₂)=0.0135 mol → n(OH^–)=2·0.0135= 0.027 mol
       · Neutralization reaction: 0.002 mol H and 0.027 mol OH^–
       · remaining 0.025 mol OH^–
       · Total volume: 100 ml
       · pOH=-log(0.025/0.1)=0.6 → pH=14-0.6=13.4

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)

Calculate the pH of the following mixtures.

1. For 80 ml of a 0.05 M solution of hydrochloric acid, 100 ml of a 0.01 M solution of sodium nitrate is given.
2. For 60 ml of a 0.015 KOH solution, 30 ml of 0.2 M saline solution is given.
3. 10 ml 0.2 M saltic acid solution is first diluted tenfold and then replaced with 1.0 g Ca(OH)₂.

1. Start- pH: pH=-log(0.03) = 1.5
       A hundred times dilution means that 99 parts of a solution are water.
       z.B. 1 ml the 0.03 M solution [translation unavailable] 99 ml water
       · 1000 ml ≙ 0.03 mol H⁺
       · 1 ml → 0.00003 mol H⁺
       · pH=-log(0.00003/0.001)=1.52
       · Note that 99 ml of water is allowed
       · pH=-log(0.00003/(0.001+0.099)) = 3.52
       It should be noted that the increase of two decimal places in
       H⁺ Concentration changes whether by 2 units at pH
2. Start- pH: pH=-log(0.05) = 1.3
       · 20 ml 0.05 M HCl is called:
       · 1000 ml ≙ 0.05 mol H⁺
       · 20 ml → 0.001 mol H⁺
       · Note that 100 ml of water is allowed
       · the new pH: pH=log(n/V)=-log(0.001/(0.02+0.1))= 2.079
3. Start pH: pOH=-log(0.2)=0.7, pH=14-0.7=13.3
       · 1000 ml ≙ 0.2 mol OH^–
       · 50 ml → 0.01 mol OH^–
       · Note that 100 ml of water is allowed
       · pOH=-log(0.01/(0.05+0.1))=1.18
       · pH=14-1.18=12.82

Form the neutralization reactions in ion type and form the corresponding salts, analogous to exercise a).

With pH = -log(c(H₃O⁺)) following:: c(H₃O⁺) = 10^-pH
It is the same.

1. The pH = -log(c(H₃O⁺)) is as follows:
       c(H₃O⁺) = 10^-pH or c(OH^–) = 10^-pOH
       Similarly, c=n/V=m/M/V and thus:
       m = c·V·M
2. (a) m(H₃O⁺) = c·V·M=10^-6 mol/l · 1.0 l · 19 g/mol = 1.9·10^-5 grams
       m(OH^–) = 10^-8 mol/l · 1.0 l · 17 g/mol = 1.7·10^-7 grams
3. b) volumes lake Constance, coarse estimated: l= 50 km, b= 2km, h=300m
       V= 30km³=30·(1000m·1000m·1000m)=10⁹ m³ = 10¹² litres
       m(H₃O⁺) = c·V·M=10^-7 mol/l · 10¹² l · 0.019 kg/mol = 1900 kg
       m(OH^–) = 10^-7 mol/l · 10¹² l · 0.017 kg/mol = 1700 kg