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Calculate The pH-Value the following: Mixtures.
- To the 80 Milligrams One of them 0.05 M Saltic acid solution will be 100 Milligrams One of them 0.01 M Sodium alloy solution given.
- To the 60 Milligrams One of them 0.015 KOH-Lthe island will be 30 Milligrams 0.2 M Saltic acid solution given.
- 10 Milligrams 0.2 M Saltic acid solution will be first. 10 times diluted and and then with
1.0 g Ca(OH)2 of which:.
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 Staying In the case of The neutralization reaction 0.003 mol HCl
· Please note:: total Volumes 80Milligrams + 100 Milligrams
· pH=-log(0.003/0.18) = 1.78 - · 60 Milligrams 0.015 M KOH → n(KOH)=0.0009 mol
· 30 Milligrams 0.2 M HCl → n(HCl)=0.006mol
· Please note:, that KOH The Commission Similarly It is How NaOH, So what do you mean? also One base is
· remaining Staying In the case of The neutralization reaction 0.0051 mol HCl
· Please note:: total Volumes 60Milligrams + 30 Milligrams
· pH=-log(0.0051/0.09) = 1.25 - · Starting-pH: pH=-log(0.2) = 0.7
· Tenfold Dilution That's right., that to The 10 Milligrams Not yet Additional 90 Milligrams This is the. Water given will be
· 10 Milligrams 0.2 M HCl → n(HCl)=0.002 mol Responsibility. 0.002 mol
· n(HCl)=0.002 mol → 0.002 mol H+
· Total Volumes: 10ml+90ml=100 Milligrams
· pH=-log(0.002/0.1) = 1.7
· Increase to One pH-Unity power In the case of 10 times Dilution Meaning
· 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 Staying 0.025 mol OH–
· total Volumes: 100 Milligrams
· pOH=-log(0.025/0.1)=0.6 → pH=14-0.6=13.4
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