By titration, it is found that 31.7 mL of 0.145 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentration of the HCl solution.

0.184 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, the following data were obtained:

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

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

Next, the data obtained from the question. This includes:

Volume of the base, NaOH (Vb) = 31.7 mL Molarity of the base, NaOH (Mb) = 0.145 M

Volume of the acid, HCl (Va) = 25.0 mL

Molarity of the acid, HCl (Ma) =?

Finally, we shall determine the molarity of the acid (HCl) as shown below:

MaVa /MbVb = nA/nB

Ma × 25 / 0.145 × 31.7 = 1

Cross multiply

Ma × 25 = 0.145 × 31.7

Ma × 25 = 4.5965

Divide both side by 25

Ma = 4.5965 / 25

Ma = 0.184 M

Therefore, the molarity of the acid (HCl) is 0.184 M

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