What is the molarity of an HCl solution if 37.0 mL is completely titrated by 56.0 mL of an NaOH solution whose concentration is 0.250 M?

A. 0.378M C. 0.000121M
B. 8288M D. 0.165M

Option A. 0.378M

Explanation:

Data obtained from the question include:

Molarity of acid (Ma) =..?

Volume of acid (Va) = 37.0 mL

Volume of base (Vb) = 56.0 mL

Molarity of base (Mb) = 0.250 M

Next, we shall write the balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Finally, we can determine the molarity of the acid as shown below :

MaVa/MbVb = nA/nB

Ma x 37 / 0.25 x 56 = 1

Cross multiply

Ma x 37 = 0.25 x 56

Divide both side by 37

Ma = 0.25 x 56 /37

Ma = 0.378M

Therefore, the molarity of the acid, HCl is 0.378M

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d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong