Question

What is the molarity of a solution of HCl if 5.00 mL of the HCl solution is

titrated with 28.6 mL of a 0.145 M NaOH solution?

Round your answer to 3
decimal places.

HCl + NaOH → NaCl + H 2 O

Answer 1

The molarity of the HCl solution is found to be 0.830 M by using the titration method with a 0.145 M NaOH solution to neutralize 5.00 mL of HCl.

Step-by-step explanation:

The question deals with the calculation of the molarity of an HCl solution using the technique of titration with a known concentration of NaOH. In this titration, the balanced chemical reaction is:

HCl + NaOH → NaCl + H2O

Given that 28.6 mL (or 0.0286 L) of 0.145 M NaOH solution is required to neutralize 5.00 mL of HCl solution, we can calculate the number of moles of NaOH reacted:

Moles of NaOH = Volume (L) × Molarity (M) = 0.0286 L × 0.145 M = 0.004149 mol NaOH

Since the molar ratio of HCl to NaOH is 1:1, the moles of HCl will also be 0.004149. To find the molarity of the HCl solution, we use the formula:

Molarity of HCl = Moles of HCl / Volume of HCl (L) = 0.004149 mol / 0.005 L = 0.830 M

Therefore, the molarity of the HCl solution is 0.830 M, rounded to three decimal places.

Answer 2

The molarity of the HCl is 0.8924 M.

Step-by-step explanation:

Use dimensional analysis to solve the problem.

28.6 mL x (
`(1 L)/(1000 mL)`) x (
`(0.145 mole NaOH)/(1 L)`) x (
`(1 mole HCl)/(1 mole NaOH)`) = 0.004147 moles HCl

Then we use M= moles/ liter to find the molarity.

M = 0.004147 moles HCl / 0.005 L

M = 0.8294