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Calculate the pH of the following mixtures.
- 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.
- For 60 ml of a 0.015 KOH solution, 30 ml of 0.2 M saline solution is given.
- 10 ml 0.2 M saltic acid solution is first diluted tenfold and then replaced with
1.0 g Ca(OH)2.
Solution: - 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 - · 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 - · 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
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