Question

How many milliliters of a 0.249 M HCl solution are needed to neutralize 213 mL of a 0.0357 M Ba(OH)2 solution?

Answer 1

Try this equation:

M1V1=M2V2

Answer 2

M stands for molarity V stands for final. 1 is initial, 2 is final

Answer 3

Do I need to convert the mL to L?

Answer 4

not necessarily, as long as you use mL in every part of the equation, and you understand that your answer will be given in mL

Answer 5

So 0.249 = (0.0357)(213)

Answer 6

Is that right?

Answer 7

M1V1=M2V2
you're solving for V2

V2= M1V1/M2

Answer 8

So (0.0357)(213)/(.249)

Answer 9

Thats right! your answer should have 3 sig figs

Answer 10

I got 30.5

Answer 11

It keeps counting it wrong....Did I do my calculations wrong? I keep getting the same answer.

Answer 12

You should be careful when calculate it. In this problem, \(\rm Ba(OH)_2\) has two \(\rm OH^-\) while \(\rm HCl\) can form only one \(\rm H^+\)

Answer 13

So was the formula wrong?

Answer 14

No it is correct. It is formed from \(\large ~n_{H^+}=n_{OH^-}\)
\(\large ~n_{H^+}=M_1V_1\)
\(\large ~n_{OH^-}=2M_2V_2\)

Answer 15

Since \(1Ba(OH)_2\longrightarrow2OH^-\), \(n_{OH^-}=2n_{Ba(OH)_2}\)

Answer 16

So I would do (2)(.0357)(.213)/(.249)

Answer 17

Yes, is that the answer?

Answer 18

I got .0610 for the answer

Answer 19

Well .0611 since the 0 would round

Answer 20

Yes, convert it into L

Answer 21

Sorry, I mean mL

Answer 22

I got 61.0771 since I just redid it with 213 instead of .213

Answer 23

That's right. See, you don't have to convert V from mL to L

Answer 24

I see now.

Answer 25

I knew I had to with molarity type problems and thought I would have to do it for this problem too.

Answer 26

Yes, you should be careful when converting between these units. I made lots of mistakes because of them

Answer 27

Thanks for the tip

Answer 28

You are very welcome. Keep up the good work

Answer 29

Thank you for your help as well, Goku

Answer 30

Thank you. I will

Answer 31

Have a great weekend

Answer 32

You too