Reply To:Calculations of pH, dilutions (SB-49) 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:

1. 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
2. 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
3. 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