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Calculate the pH of the following solutions if the solvent is always pure water:
- A 0.03 molar HCl solution is diluted a hundred times.
Calculate the pH before and after dilution - 20 ml 0.05 M HCl is diluted with 100 ml distilled water.
Calculate the pH before and after dilution - 50 ml of 0.2 mo of NaOH solution is given 100 ml of distilled water.
Calculate the pH before and after dilution
Solution: - 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 - 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 - 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
- A 0.03 molar HCl solution is diluted a hundred times.
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