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What volume (in mL) of 8.84 M HBr would be required to make 300.0 mL of a solution with a pH of 2.99
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Farah
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Answer:
0.0347 mL of 8.84 M HBr would be required to make 300.0 mL of a solution with a pH of 2.99
Explanation:
Dilution is the reduction of the concentration of a chemical in a solution.
Dilution consists of adding more solvent. The amount of solute does not change, but the volume of the solvent does: as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.
Since the amount of solute does not vary, then it can be expressed as:
initial number of moles = final number of moles
The initial number of moles can be expressed as:
Initial concentration * volume initial
where the initial concentration is 8.84 M and the volume is the data to be calculated.
The final number of moles can be expressed as:
final concentration * volume final
pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance. PH is defined as the opposite of the base 10 logarithm or the negative logarithm of the activity of hydrogen ions:
pH= -log [H⁺]
If the pH = 2.99, then:
2.99= -log [H⁺]
1.023*10⁻³ M= [H⁺]
A strong acid is totally dissociated, which implies that the acid concentration equals the final H⁺ concentration. Then:
[H⁺]= [HBr]= 1.023*10⁻³ M
Being 300 mL= 0.300 L, you can replace in:
Final number of moles = final concentration * volume final
Final number of moles= 1.023*10⁻³ M* 0.300 L
Final number of moles= 3.069*10⁻⁴ moles
So:
initial number of moles = final number of moles
8.84 M* volume initial= 3.069*10⁻⁴ moles
volume initial= 3.069*10⁻⁴ moles ÷ 8.84 M
volume initial= 3.47* 10⁻⁵ L= 0.0347 mL
0.0347 mL of 8.84 M HBr would be required to make 300.0 mL of a solution with a pH of 2.99