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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: - 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 - · 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 - · 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
· 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
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