Calculate the conce… - QuestionCove
OpenStudy (anonymous):

Calculate the concentration (in molarity) of an NaOH solution if 25.0mL of the solution is needed to neutralize 17.9 mL of a 0.323 M HCl solution.

2 years ago
OpenStudy (anonymous):

Ok, so we need to calculate the concentration, right? What is the formula?

2 years ago
OpenStudy (anonymous):

C = (# moles)/(vol of solution in liters)

2 years ago
OpenStudy (anonymous):

That's right. And we are given...?

2 years ago
OpenStudy (anonymous):

volume solution =25.0

2 years ago
OpenStudy (anonymous):

Perfect. That leaves us moles to find

2 years ago
OpenStudy (anonymous):

Yes

2 years ago
OpenStudy (anonymous):

How would you find it with these information? "the solution is needed to neutralize 17.9 mL of a 0.323 M HCl solution."

2 years ago
OpenStudy (anonymous):

Getting moles from the molar mass of NaOH

2 years ago
OpenStudy (anonymous):

No that wouldn't be right

2 years ago
OpenStudy (anonymous):

You are given C and V of HCl. What can you get from these info?

2 years ago
OpenStudy (anonymous):

Would you multiple the (17.9)(.323)

2 years ago
OpenStudy (anonymous):

Right. That's moles of HCl, right?

2 years ago
OpenStudy (anonymous):

yes

2 years ago
OpenStudy (anonymous):

=5.7817 is what I got

2 years ago
OpenStudy (anonymous):

Yes, and the unit is mmol since you haven't convert mL to L. But you don't have to

2 years ago
OpenStudy (anonymous):

Ok. Now, how that's gonna help us find the moles of NaOH that we need?

2 years ago
OpenStudy (anonymous):

M = (5.7817)/(.025) =

2 years ago
OpenStudy (anonymous):

Because H = OH

2 years ago
OpenStudy (anonymous):

That's right.

2 years ago
OpenStudy (anonymous):

However, since it is 5.7817 mmol, the volume should be in mL

2 years ago
OpenStudy (anonymous):

Because you haven't converted 17.9 mL to L

2 years ago
OpenStudy (anonymous):

Right

2 years ago
OpenStudy (anonymous):

.231268

2 years ago
OpenStudy (anonymous):

2 years ago
OpenStudy (anonymous):

It's much better than before. That's for sure

2 years ago
OpenStudy (anonymous):

I get tripped up on the problems if they are worried differently

2 years ago
OpenStudy (anonymous):

See, it is like you are reading a question and you search an article to find the answer

2 years ago
OpenStudy (anonymous):

You look for what you need to help answer it

2 years ago
OpenStudy (anonymous):

Since you know what you need to find, different info won't affect much

2 years ago
OpenStudy (anonymous):

Do you have any more problems?

2 years ago
OpenStudy (anonymous):

I want to see how you plan to solve it

2 years ago
OpenStudy (anonymous):

I do

2 years ago
OpenStudy (anonymous):

2 years ago
OpenStudy (anonymous):

Btw, well done

2 years ago
OpenStudy (anonymous):

Thank you. You are good at explaining

2 years ago
OpenStudy (anonymous):

Thank you

2 years ago
OpenStudy (anonymous):

You're welcome

2 years ago
OpenStudy (anonymous):

What volume of a 5.00 M HCl solution is needed to neutralize each of the following: (a) 20.0 mL of a 0.300 M NaOH solution (b) 17.0 mL of a 0.200 M Ba(OH)2 solution

2 years ago
OpenStudy (anonymous):

Ok. What is your thought? What should we do first?

2 years ago
OpenStudy (anonymous):

Try to find the volume since that is what it is asking for

2 years ago
OpenStudy (anonymous):

That's right. Then how do you do it?

2 years ago
OpenStudy (anonymous):

Is this one going to be like the one we just did?

2 years ago
OpenStudy (anonymous):

Exactly the same. Both acid and base are monoprotic

2 years ago
OpenStudy (anonymous):

However, for (b), you should pay attention to the OH- group. $$1Ba(OH)_2\longrightarrow 2OH^-$$

2 years ago
OpenStudy (anonymous):

So we need to find moles?

2 years ago
OpenStudy (anonymous):

Yes

2 years ago
OpenStudy (anonymous):

So you would find moles of the NaOH

2 years ago
OpenStudy (anonymous):

That's correct. Then from there, you can calculate V of HCl

2 years ago
OpenStudy (anonymous):

So you would use the .300 to get the moles right?

2 years ago
OpenStudy (anonymous):

Yes

2 years ago
OpenStudy (anonymous):

.300/39.988= .0075022507 moles

2 years ago
OpenStudy (anonymous):

No. Moles of NaOH = concentration x volume

2 years ago
OpenStudy (anonymous):

11.9964

2 years ago
OpenStudy (anonymous):

You can't divide the concentration by MM of NaOH to find moles, right?

2 years ago
OpenStudy (anonymous):

Yeah

2 years ago
OpenStudy (anonymous):

Wait, are you on (a) or (b)? Because they are different problems

2 years ago
OpenStudy (anonymous):

a

2 years ago
OpenStudy (anonymous):

Alright. How did you calculate moles of NaOH?

2 years ago
OpenStudy (anonymous):

.300 * 39.988 = 11.9964

2 years ago
OpenStudy (anonymous):

Wait. Moles of NaOH = concentration x volume = .300M x .0200L, right? We are not supposed to use MM

2 years ago
OpenStudy (anonymous):

Oh

2 years ago
OpenStudy (anonymous):

I got .006

2 years ago
OpenStudy (anonymous):

That's right. Now you can calculate C of HCl

2 years ago
OpenStudy (anonymous):

Find moles and use the .0200 L

2 years ago
OpenStudy (anonymous):

.500 * .0200 = .01

2 years ago
OpenStudy (anonymous):

Moles of NaOH = concentration x volume => Concentration = Moles of NaOH / volume

2 years ago
OpenStudy (anonymous):

15

2 years ago
OpenStudy (anonymous):

(.300)/(.0200)

2 years ago
OpenStudy (anonymous):

No, you already got the moles here "I got .006". So it should be .006moles/5.000M

2 years ago
OpenStudy (anonymous):

Oh

2 years ago
OpenStudy (anonymous):

1.2

2 years ago
OpenStudy (anonymous):

That's right

2 years ago
OpenStudy (anonymous):

Are you tired?

2 years ago
OpenStudy (anonymous):

Somewhat. It's been a long day but I wanna finish the homework tonight

2 years ago
OpenStudy (anonymous):

Ok. Then let's move on to (b)

2 years ago
OpenStudy (anonymous):

The 1.2 is the answer to a right?

2 years ago
OpenStudy (anonymous):

That's right. 1.2mL. Oops, I thought we are calculating the concentration. So I have been confusing you. I am so sorry

2 years ago
OpenStudy (anonymous):

It's okay. You didn't mean to

2 years ago
OpenStudy (anonymous):

We're probably both tired

2 years ago
OpenStudy (anonymous):

Ok I will do my best on (b)

2 years ago
OpenStudy (anonymous):

First let's calculate $$n_{OH-}$$

2 years ago
OpenStudy (anonymous):

Are you awake?

2 years ago
OpenStudy (anonymous):

.0034

2 years ago
OpenStudy (anonymous):

Yeah. Just took me a bit to get it

2 years ago
OpenStudy (anonymous):

Lol. Just to make sure

2 years ago
OpenStudy (anonymous):

That's moles of Ba(OH)2?

2 years ago
OpenStudy (anonymous):

Yes

2 years ago
OpenStudy (anonymous):

Ok. What is the moles of HCl reacted with that amount of Ba(OH)2?

2 years ago
OpenStudy (anonymous):

.0017

2 years ago
OpenStudy (anonymous):

Wait you divide so .0068

2 years ago
OpenStudy (anonymous):

That's right this time

2 years ago
OpenStudy (anonymous):

The .0017?

2 years ago
OpenStudy (anonymous):

Good job. Don't make mistakes on the easy ones :)

2 years ago
OpenStudy (anonymous):

No. .0068 is correct

2 years ago
OpenStudy (anonymous):

Good

2 years ago
OpenStudy (anonymous):

Now could you tell me the volume of HCl needed?

2 years ago
OpenStudy (anonymous):

Would you take (.500)(.0068)

2 years ago
OpenStudy (anonymous):

No. V = moles / C = .0068 / 5.000

2 years ago