what is the molarity of an HCI solution if 25.0 ml of 0.185 M NaOH is required to neutralize 0.0200 L of HCI?

0.231 M.

Explanation:

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

HCl + NaOH —> NaCl + H2O

From the balanced equation above, we obtained the following data:

Mole ratio of the acid, HCl (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, the data obtained from the question:

Volume of base, NaOH (Vb) = 25 mL

Molarity of base, NaOH (Mb) = 0.185 M

Volume of acid, HCl (Va) = 0.0200 L

Molarity of acid, HCl (Ma) =?

Next, we shall convert 0.0200 L to mL

This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.0200 L = 0.0200 L × 1000 mL /1 L

0.0200 L = 20 mL

Thus, 0.0200 L is equivalent to 20 mL.

Finally, we shall determine the molarity of the acid solution as follow:

Mole ratio of the acid, HCl (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Volume of base, NaOH (Vb) = 25 mL

Molarity of base, NaOH (Mb) = 0.185 M

Volume of acid, HCl (Va) = 20 mL

Molarity of acid, HCl (Ma) =?

MaVa / MbVb = nA/nB

Ma × 20 / 0.185 × 25 = 1

Ma × 20 / 4.625 = 1

Cross multiply

Ma × 20 = 4.625 × 1

Ma × 20 = 4.625

Divide both side by 20

Ma = 4.625 / 20

Ma = 0.231 M

Thus, the molarity of the acid solution is 0.231 M.

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