How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

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How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

Engineering

Chemistry

Dilution and Mixing of Solution

pH Calculation for Acids and Bases

Question

How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

A

0.1 L

B

9.0 L

C

2.0 L

D

0.9 L

10^–1 × 1 = 10^–2 × V_f

$∴$ V_f = 10 L

$∴$ Volume of water (V_f – V_i) = 9 L

Concept Explanation: Dilution of Acid Solution

This problem involves diluting an acidic solution to change its pH. pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration, [H⁺]. The formula is:

$p H = - \log [H^{+}]$

When we dilute an acid by adding water, the number of moles of H⁺ ions remains constant, but their concentration decreases because the total volume increases.

Step-by-Step Solution:

Step 1: Find the initial [H⁺] concentration.

Initial pH = 1

Using the pH formula: $p H = - \log [H^{+}]$

Therefore, $1 = - \log [H^{+}]$

This means $\log [H^{+}] = - 1$

So, the initial concentration, [H⁺]_initial = 10^-1 M = 0.1 M

Step 2: Find the final [H⁺] concentration.

Final pH = 2

Using the same formula: $2 = - \log [H^{+}]$

Therefore, $\log [H^{+}] = - 2$

So, the final concentration, [H⁺]_final = 10^-2 M = 0.01 M

Step 3: Apply the dilution principle.

The number of moles of H⁺ ions remains unchanged during dilution. The formula for dilution is:

$M_{1} V_{1} = M_{2} V_{2}$

Where:
M₁ = Initial Molarity = 0.1 M
V₁ = Initial Volume = 1 L
M₂ = Final Molarity = 0.01 M
V₂ = Final Volume = ? L

Plugging the values into the equation:

$(0.1) (1) = (0.01) (V_{2})$

$0.1 = 0.01 V_{2}$

Solving for V₂:

$V_{2} = \frac{0.1}{0.01} = 10 L$

Step 4: Calculate the volume of water added.

The final volume (V₂) is the sum of the original solution and the water added.

Let the volume of water added be $x$ litres.

Therefore: Final Volume = Initial Volume + Water Added

$10 = 1 + x$

Solving for x:

$x = 10 - 1 = 9$

Final Answer: You must add 9.0 L of water.

Related Topics & Formulae:

1. pH Scale: A scale from 0 to 14 used to specify the acidity or basicity of an aqueous solution. It is a logarithmic scale.

Key Formula: $p H = - \log [H^{+}]$

2. Dilution: The process of reducing the concentration of a solute in a solution, usually by adding more solvent (e.g., water).

Key Formula: $M_{1} V_{1} = M_{2} V_{2}$

Where M₁ and V₁ are the molarity and volume before dilution, and M₂ and V₂ are the molarity and volume after dilution.

3. Strong Acids: Acids that completely dissociate in water (like HCl in this problem). For strong monoprotic acids, [H⁺] equals the initial concentration of the acid.