What volume (in mL) of 8.84 M HBr would be required to make 300.0 mL of a solution with a pH of 3.15?

The volume of 8.84 M HBr needed to prepare 300.0 mL of a solution with a pH of 3.15 is 0.024 mL

We'll begin by calculating the concentration of Hydrogen ion in the resulting solution. This can be obtained as:

pH = 3.15

pH = –Log [H⁺]

3.15 = –Log [H⁺]

–3.15 = Log [H⁺]

Take the antilog of (–3.15)

[H⁺] = antilog of (–3.15)

HBr(aq) —> H⁺(aq) + Br¯(aq)

From the balanced equation above,

1 mole of HBr contains 1 mole of H⁺

Therefore,

7.08×10¯⁴ M HBr will also contain 7.08×10¯⁴ M H⁺

Molarity of stock solution (M₁) = 8.84 M

Volume of diluted solution (V₂) = 300 mL

Molarity of diluted solution (M₂) = 7.08×10¯⁴ M

8.84 × V₁ = 7.08×10¯⁴ × 300

8.84 × V₁ = 0.2124

Divide both side by 8.84

V₁ = 0.2124 / 8.84

Therefore, the volume of 8.84 M HBr needed to prepare the solution is 0.024 mL

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Volume (in mL) of 8.84 M HBr : 0.024 ml

pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.  

pH = - log [H⁺]  

pH=3.15

Volume of solution : 300 ml=0.3 L

mol of solution = M x V

Molarity of HBr = 8.84 M

The volume of HBr = mol : molarity

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