130k views
1 vote
What is the molarity of a solution of NaOH if 25 mL of 1,2 M HCI is required to neutralize 15 mL of the base?

User DavB
by
8.3k points

1 Answer

2 votes

Answer:

We can use the balanced chemical equation for the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH):

HCl + NaOH → NaCl + H2O

From the balanced equation, we can see that the mole ratio of HCl to NaOH is 1:1. That means that the number of moles of HCl that react with NaOH is equal to the number of moles of NaOH present in the solution.

We can use the equation:

Molarity = moles of solute / volume of solution (in liters)

to find the molarity of the NaOH solution. First, we need to calculate the number of moles of HCl that reacted with the NaOH. We can do this using the molarity and volume of the HCl solution:

moles of HCl = molarity x volume (in liters)

moles of HCl = 1.2 mol/L x 0.025 L

moles of HCl = 0.03 mol

Since the mole ratio of HCl to NaOH is 1:1, we know that 0.03 moles of NaOH also reacted. Now we can calculate the molarity of the NaOH solution using the volume of the NaOH solution that was used to neutralize the HCl:

Molarity of NaOH = moles of NaOH / volume of NaOH solution (in liters)

Molarity of NaOH = 0.03 mol / 0.015 L

Molarity of NaOH = 2 M

Therefore, the molarity of the NaOH solution is 2 M.

Step-by-step explanation:

User Oskar Dajnowicz
by
8.5k points

Related questions

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.