Reply To:Calculations of pH, mixtures (SB-61) Calculate the pH of the following mixtures. To the 80 ml of one 0.05 M saltic acid solution 100 ml of one 0.01 I've given you the sodium salt solution. To the 60 ml of one 0.015 KOH- Become a solution 30 Milligrams 0.2 M saltic acid solution given. 10 Milligrams 0.2 M saltic acid solution is first diluted ten times and then with nobr1.0 g} Ca(OH)2 It's been shifted.

Lösung:

1. c=n/V → n=c·V
       · 80 Milligrams 0.05 M HCl → n(HCl)=0.004 mol
       · 100 Milligrams 0.01 NaOH → n(NaOH)=0.001 mol
       · remaining in the neutralisation reaction 0.003 mol HCl
       · Note: total volume is 80ml + 100 Milligrams
       · pH = log0.003/0.18) = 1.78
2. · 60 Milligrams 0.015 M KOH → n(KOH)=0.0009 mol
       · 30 Milligrams 0.2 M HCl → n(HCl)=0.006 mol
       · Please note that KOH It behaves similarly to NaOHSo also a base is
       · remaining in the neutralisation reaction 0.0051 mol HCl
       · Please note: total volume 60ml + 30 Milligrams
       · pH = log0.0051/0.09) = 1.25
3. · Starting pH: pH = log0.2) = 0.7
       · Tenfold dilution means that to the 10 ml or more 90 ml of water is given
       · 10 Milligrams 0.2 M HCl → n(HCl)=0.002 mol The Commission shall adopt delegated acts in accordance with this Article. 0.002 mol
       · n(HCl)=0.002 mol → 0.002 mol H⁺
       · Total volume: 10ml+90ml=100 Milligrams
       · pH = log0.002/0.1) = 1.7
       · Increase by one unit of pH makes sense at ten times dilution
      
   

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       · 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^–
       · the total volume: 100 Milligrams
       · pOH=-log0.025/0.1)=0.6 → pH =14-0.6=13.4