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Answer: The molarity of an HCl solution is 0.218 M if 43.6 mL of a 0.125 M NaOH solution are needed to titrate a 25.0 mL sample of the acid.
Explanation:
Given: [tex]V_{1}[/tex] = 43.6 mL, Â Â Â [tex]M_{1}[/tex] = 0.125 M
[tex]V_{2}[/tex] = 25.0 mL, Â Â Â Â [tex]M_{2}[/tex] = ?
Formula used to calculate the concentration of acid is as follows.
[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]
Substitute the values into above formula.
[tex]M_{1}V_{1} = M_{2}V_{2}\\0.125 M \times 43.6 mL = M_{2} \times 25.0 mL\\M_{2} = 0.218 M[/tex]
Thus, we can conclude that the molarity of an HCl solution is 0.218 M if 43.6 mL of a 0.125 M NaOH solution are needed to titrate a 25.0 mL sample of the acid.
The molarity of the HCl solution used in the neutralization reaction has been 0.218 M.
The reaction of NaOH with HCl has been a neutralization reaction. The resultant will be salt and water.
In the reaction the molarity can be calculated as:
Molarity of HCl [tex]\rm[/tex][tex]\times[/tex] Volume of HCl = Molarity of NaOH
Given,
The volume of HCl solution = 25 ml
Molarity of NaOH = 0.125 M
Volume of NaOH solution = 43.6 ml
Substitute the values in the equation:
Molarity of HCl [tex]\times[/tex] 25 = 0.125 [tex]\times[/tex] 43.6
Molarity of HCl [tex]\times[/tex] 25 = 5.45
Molarity of HCl solution = [tex]\rm \dfrac{5.45}{25}[/tex]
Molarity of HCl solution = 0.218 M
The molarity of the HCl solution used in the neutralization reaction has been 0.218 M.
For more information about the molarity, refer to the link:
https://brainly.com/question/12127540
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