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What is the molarity of an HCl solution if 43.6 mL of a 0.125 M NaOH solution are needed to titrate a 25.0 mL sample of the acid according to the equation below (show your calculations)? NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)

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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,      [tex]M_{1}[/tex] = 0.125 M

[tex]V_{2}[/tex] = 25.0 mL,        [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:

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