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enot [183]
3 years ago
13

A 25.00-mL aliquot of an unstandardized HCl solution is titrated with the previously standardized NaOH solution from #1 above. I

f 32.55 mL of NaOH titrant is required to reach the endpoint, what is the exact molarity of the HCl solution?
Chemistry
1 answer:
jeka57 [31]3 years ago
5 0
The neutralization reaction would be:

HCl + NaOH --> NaCl + H₂O

Next, determine the equivalent moles of HCl from the given amount of titrant. However, there is a missing information. You should know the molarity of the NaOH to be able to solve this. Suppose, the concentration is 0.1 M NaOH.

0.1 M NaOH * 0.3255 L = 0.03255 mol NaOH

Since, the molar ratio is 1:1, then that is also equivalent to 0.03255 mol HCl. The molarity would then be:

Molarity = 0.03255 mol/0.25 = <em>0.1302 M HCl</em>
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The molarity of the HCl solution needed to neutralize 28.6 mL of a 0.175 M NaOH solution is 0.2002 M

We'll begin by writing the balanced equation for the reaction. This is given below:

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From the balanced equation above,

The mole ratio of the acid, HCl (nA) = 1

The mole ratio of the base, NaOH (nB) = 1

  • From the question given above, the following data were obtained:

Volume of base, NaOH (Vb) = 28.6 mL

Molarity of base, NaOH (Mb) = 0.175 M

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<h3>Molarity of acid, HCl (Ma) = ?</h3>

The molarity of the acid, HCl can be obtained as follow:

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Cross multiply

Ma × 25 = 5.005

Divide both side by 25

Ma = 5.005 / 25

<h3>Ma = 0.2002 M</h3>

Therefore, the molarity of the acid, HCl needed for the reaction is 0.2002 M

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