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Calculate The pH-Values the following: LManufacture in which all the materials used are:, This LVegetable products be Always of a purity Water:
- One of them. 0.03 molar HCl-Lthe island will be One hundred times diluted.
Calculate The pH before and after The Dilution - 20 Milligrams 0.05 M HCl will be with 100 Milligrams Distilled Water diluted.
Calculate The pH before and after The Dilution - To the 50 Milligrams One of them 0.2 molars NaOH-Lthe island will be 100 Milligrams Distilled Water given.
Calculate The pH before and after The Dilution
Lösung: - Starting-pH: pH=-log(0.03) = 1.5
One of them. One hundred times Dilution That's right., that on a Part of Lthe island 99 Parts and accessories Water Come on .
z.B. 1 Milligrams The 0.03 M Lthe island plus 99 Milligrams Water
· 1000 Milligrams ≙ 0.03 mol H+
· 1 Milligrams → 0.00003 mol H+
· pH=-log(0.00003/0.001)=1.52
· Please note:, that 99 Milligrams Water Admitted will be
· pH=-log(0.00003/(0.001+0.099)) = 3.52
Please note:, that The Commission The Increase to two 10th power with regard to: The
H+ Concentration Changes, whether or not to 2 Units At the same time, pH-Value - Starting-pH: pH=-log(0.05) = 1.3
· 20 Milligrams 0.05 M HCl heated:
· 1000 Milligrams ≙ 0.05 mol H+
· 20 Milligrams → 0.001 mol H+
· Please note:, that 100 Milligrams Water Admitted will be
· new pH: pH=log(n/V)=-log(0.001/(0.02+0.1))= 2.079 - Starting-pH: pOH=-log(0.2)=0.7, pH=14-0.7= 13.3
· 1000 Milligrams ≙ 0.2 mol OH–
· 50 Milligrams → 0.01 mol OH–
· Please note:, that 100 Milligrams Water Admitted will be
· pOH=-log(0.01/(0.05+0.1))=1.18
· pH=14-1.18=12.82
- One of them. 0.03 molar HCl-Lthe island will be One hundred times diluted.
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