Question

If 3.00 liters of 2M HCl neutralizes 1.00 liter of NaOH, what is the molarity of the NaOH?

Answer 1

`M_(base)=6M`

Step-by-step explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize this is a question about titration, which base solved by knowing that the HCl reacts with the NaOH in a 1:1 mole ratio, and therefore, we can write the following:

`M_(acid)V_(acid)=M_(base)V_(base)`

Thus, we solve for the molarity of the base, NaOH, as shown below:

`M_(base)=(M_(acid)V_(acid))/(V_(base)) \\\\M_(base)=(3L*2M)/(1.00L)\\\\M_(base)=6M`

Regards!