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Calculate the pH of the following solutions if the solvent is always pure water:
- One of them. 0.03 molar HCl- The solution is diluted a hundred times.
Calculate the pH before and after dilution - 20 Milligrams 0.05 M HCl will be 100 diluted in ml of distilled water.
Calculate the pH before and after dilution - To the 50 ml of one 0.2 molars NaOH- The solution will be 100 ml of distilled water.
Calculate the pH before and after dilution
Lösung: - Starting pH: pH = log0.03) = 1.5
A hundred-fold dilution means that on a partial solution 99 Parts of water coming
z.B. 1 ml of 0.03 M solution plus 99 ml of water
· 1000 Milligrams ≙ 0.03 mol H+
· 1 Milligrams → 0.00003 mol H+
· pH = log0.00003/0.001)=1.52
· Please note that 99 ml of water to be admitted
· pH = log0.00003/(0.001+0.099)) = 3.52
note that the increase by two decimal places in relation to the
H+ Concentration changes whether to 2 Units of pH - Starting pH: pH = log0.05) = 1.3
· 20 Milligrams 0.05 M HCl the name:
· 1000 Milligrams ≙ 0.05 mol H+
· 20 Milligrams → 0.001 mol H+
· Please note that 100 ml of water to be admitted
· The new pH: pH=log(n/V)=-log0.001/(0.02+0.1))= 2.079 - Start pH: pOH=-log0.2)=0.7, pH=14-0.7= 13.3
· 1000 Milligrams ≙ 0.2 mol OH–
· 50 Milligrams → 0.01 mol OH–
· Please note that 100 ml of water to be admitted
· pOH=-log0.01/(0.05+0.1))=1.18
· pH =14-1.18=12.82
- One of them. 0.03 molar HCl- The solution is diluted a hundred times.
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