The following data were collected at the endpoint of a titration performed to find the molarity of an HCl solution:

Volume of acid (HCl) used = 14.4 mL; Volume of base (NaOH) used = 22.4 mL;
Molarity of standard base (NaOH) = 0.200 M;
What is the molarity of the acid solution?

A. 1.6 M
B. 0.64 M
C. 0.31M
D. 0.13 M

C.0.31M

To find the molarity of the acid solution (HCl), you can use the concept of stoichiometry and the balanced chemical equation for the reaction between HCl and NaOH. The balanced chemical equation is:

HCl + NaOH → NaCl + H2O

From the equation, you can see that the mole ratio between HCl and NaOH is 1:1. This means that the moles of HCl is equal to the moles of NaOH used in the titration.

First, calculate the moles of NaOH used:
moles of NaOH = Molarity of NaOH × Volume of NaOH used (in liters)
= 0.200 M × 0.0224 L
= 0.00448 moles

Since the mole ratio between HCl and NaOH is 1:1, the moles of HCl is also 0.00448 moles.

Next, determine the molarity of the HCl solution:
Molarity of HCl = Moles of HCl / Volume of HCl used (in liters)
= 0.00448 moles / 0.0144 L
≈ 0.311 M

Therefore, the molarity of the acid solution (HCl) is approximately 0.311 M.

So, the correct answer is C. 0.31 M.

In 1:1 reactions as one has in HCl and NaOH, the simplest formula to use is

M x mL = M x mL.

The answer is C.0.13