Question

how much water should be added to 1L of an aqueos solution of hcl having pH=1
to prepare a solution having pH=2

Answer 1

Explanation:

pH is defined as the negative logarithm of the hydrogen concentration of an aqueous solution .

In the first case , the pH of the solution is 1 .

1 = - log₁₀ H+

⇒ 1 = log₁₀ ( 1/H ) [ - log H = log ( 1/H ) ]

⇒ 1/H = 10¹

⇒ H = 1/10

⇒ H = 0.1 .

So the hydrogen concentration is 0.1 .

For the second case :

2 = - log₁₀ H+

⇒ 2 = log₁₀ ( 1/H )

⇒ 1/H = 10²

⇒ H = 1/100

⇒ H = 0.01 .

So the hydrogen concentration is 0.01 .

Originally there was 1 L of water . Let V litres of water be added to the solution . So the volume will become V + 1 .

M₁V₁ = M₂V₂ [ M represents molarity while V is volume ]

⇒ 0.1 × 1 = 0.01 × ( 1 + V )

⇒ 0.1 = 0.01 ( 1 + V )

⇒ 10 = 1 ( 1 + V )

⇒ 10 = 1 + V

⇒ V = 10 - 1

⇒ V = 9 .

The volume of water that needs to be added is 9 L .