Used 24.13mL of 0.111M sodium Hydroxide to neutralize 20.00mL of hydrochloric acid solution.

Question

What is the molarity of the acid solution?

Answer 1

What is the molarity of the HCL solution before you add the NaOH? This is necessary to solve the problem

Answer 2

Molarity is the number of moles present in a 1L solution.

You have 24.13 mL sample of a 0.111M NaOH solution, or a 24.13 mL sample from 1L which contains 0.111 moles.

Using the formula concentration= moles/volume, we get

0.111=(moles/24.13)*1000; <volume is in mL, not L>

Therefore you have 24.13*0.111/1000=0.00267843 moles of NaOH.

Now you need to form a balanced equation.

Acid+base-->Salt + water, so

HCl + NaOH --> NaCl + H2O.

Now look at the ratio of moles in the equation of NaOH and HCl, they are in a 1:1 ratio, that is 1 mole of NaOH reacts with 1 mole of HCl.

So moles of HCl are the same, that is 0.00267843.

Now, volume= 20, and you have moles too, plug it in to the formula,

concentration= 0.00267843/20 * 1000= 0.134M (3 significant figures)

                    

Answer 3

24.13 mL x 0.111 M = 20.00 mL x HCl M

HCl = 0.134 M