Question

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

Answer 1

Answer:

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

Answer 2

Answer:

A. 0.378M

Explanation:

Hello,

In this case, by knowing the neutralization acid-base reaction between hydrochloric acid and sodium hydroxide:

$NaOH+HCl\rightarrow NaCl+H_2O$

In a titration process, at the equivalence point we notice that the moles of acid equal the moles of base:

$n_{HCl}=n_{NaOH}$

That in terms of molarities and volumes turns out:

$M_{HCl}V_{HCl}=M_{NaOH}V_{NaOH}$

Hence, with the given information we are asked to compute the molarity of the acid:

$M_{HCl}=\frac{M_{NaOH}V_{NaOH}}{V_{HCl}} =\frac{0.250M*56.0mL}{37.0mL} \\\\M_{HCl}=0.378M$

Thereby, answer is A. 0.378M.

Best regards.