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What volume (in mL) of 8.84 M HBr would be required to make 300.0mL of a solution with a pH of 3.43?

User Ashgromnies
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15 votes

So,

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

Replacing our value for pH, we're going to solve for [H3O+]:

Next, we shall determine the molarity of the resulting solution of HBr.

When HBr dissociates, the following reaction takes place:

From the balanced equation above, we can notice that 1 mole of HBr contains 1 mole of H⁺

Therefore, 3.71×10¯⁴ M HBr will also contain 3.71×10¯⁴ M H⁺.

Finally, we shall determine the volume of the stock solution of HBr needed to prepare the solution. From the values of the problem, we know that:

Molarity of stock solution (M₁) = 8.84 M

Volume of diluted solution (V₂) = 300 mL

Molarity of diluted solution (M₂) = 3.71×10¯⁴ M

Volume of stock solution needed (V₁) = ?

Using the following equation, we could solve for V1:

Therefore, 0.013 mL of 8.84M HBr would be needed.

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User John C Earls
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