A solution was prepared as follows : 100ml of 0.1m hcl + 300ml of 0.1m h2so4 + 100ml of 0.3m ba(oh)2 and volume was made to 1l by adding water. The ph of the solution would be :
Answers
The pH of the solution will be 2
Explanation:
H₂SO₄ and ba(oh)₂ present in the solution will react completely to form salt and water. The reaction is given below:
H₂SO₄ + Ba(OH)₂ = BaSO₄ + 2H₂O
The molar concentration of 300 ml of 0.1m H₂SO₄ = 300 x 0.1 = 30m
The molar concentration of 100 ml of 0.3m Ba(OH)₂ = 100 x 0.3 = 30m
Hence they will neutralize each other completely. The only PH will be of hcl.
Total volume of the solution after adding water = 1000 ml
=> concentration of Hcl after adding water = 0.1 x (100/1000) = 0.01
Hence,
pH = - log[Hcl]
= - log[0.01]
= 2
Hence the pH of the solution will be 2
The pH of the solution will be 2
Explanation:
H₂SO₄ and Ba(oh)₂ present in the solution will react completely to form salt and water. The reaction is given below:
H₂SO₄ + Ba(OH)₂ = Ba SO₄ + 2H₂O
The molar concentration of 300 ml of 0.1m H₂SO₄ = 300 x 0.1 = 30m
The molar concentration of 100 ml of 0.3m Ba(OH)₂ = 100 x 0.3 = 30m
Hence they will neutralize each other completely. The only PH will be of hcl.
Total volume of the solution after adding water = 1000 ml
=> concentration of H Cl after adding water = 0.1 x (100/1000) = 0.01
Hence,
pH = - log[H cl]
= - log[0.01]
= 2
Hence the pH of the solution will be 2
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