## CHEMISTRY MOLARITY QUESTION

**Moderators:** Leonid, amiradm, BioTeam

### CHEMISTRY MOLARITY QUESTION

Hey guys, I am trying to study for a chemistry final and I am struggling, even on the basics.

I was wondering if one of you guys could help me out if it isn't asking for too much.

Here is the question:

A solution of HCL is prepared by diluting 0.500L of HCL solution to 6.0 L. If 10.0 mL of the diluted HCL solution is required to titrate 20.0mL of 0.25 mol/L NaOH solution, then what is the concentration of the undiluted HCL solution?

This is what I thought I could do:

Find the amount of moles of NaOH by using the Molarity equation.

Since this is a titration and we are dealing with a strong base and a strong acid, for the solution to neutralize we need the same amount of moles of both.

Since there are 0.005 mol of NaOH I thought (what seems logical) that if I plug that amount of moles into the molarity equation of HCL I should get the right answer. But I am not able to come up with the right answer, so clearly I am missing something.

Care to explain guys?

I am so useless at chem which is scary since I have to take 3 more chem courses to complete my degree.

I was wondering if one of you guys could help me out if it isn't asking for too much.

Here is the question:

A solution of HCL is prepared by diluting 0.500L of HCL solution to 6.0 L. If 10.0 mL of the diluted HCL solution is required to titrate 20.0mL of 0.25 mol/L NaOH solution, then what is the concentration of the undiluted HCL solution?

This is what I thought I could do:

Find the amount of moles of NaOH by using the Molarity equation.

Since this is a titration and we are dealing with a strong base and a strong acid, for the solution to neutralize we need the same amount of moles of both.

Since there are 0.005 mol of NaOH I thought (what seems logical) that if I plug that amount of moles into the molarity equation of HCL I should get the right answer. But I am not able to come up with the right answer, so clearly I am missing something.

Care to explain guys?

I am so useless at chem which is scary since I have to take 3 more chem courses to complete my degree.

### Re: CHEMISTRY MOLARITY QUESTION

Let's see if I can do this.

Ok, first write the balanced equation. NaOH+HCL---> NaCl + H2O

So we see from the equation that it is a 1 to 1 mole ratio.

Since you determined that .005 moles of NaOH were titrated. You now know that .005 mole HCL was used. Molarity of HCL = .005mol/.01L = .5M

Our titrating HCL had a molar of .5M

Now using the formula M1V1=M2V2 We have M1(.5L)=.5M(6L)

transposing we get M1= (.5Mx6L)/.5L = 6M

undiluted HCL = 6M

Hows that sound?

I assume you have the right answer to double check? I hope...

Ok, first write the balanced equation. NaOH+HCL---> NaCl + H2O

So we see from the equation that it is a 1 to 1 mole ratio.

Since you determined that .005 moles of NaOH were titrated. You now know that .005 mole HCL was used. Molarity of HCL = .005mol/.01L = .5M

Our titrating HCL had a molar of .5M

Now using the formula M1V1=M2V2 We have M1(.5L)=.5M(6L)

transposing we get M1= (.5Mx6L)/.5L = 6M

undiluted HCL = 6M

Hows that sound?

I assume you have the right answer to double check? I hope...

### Re: CHEMISTRY MOLARITY QUESTION

Glad to be of help. Just remember to write the balanced equation first. For the question on your

H2SO4 + 2NaOH ----> Na2SO4 + 2H2O

From the equation above you can see that 1 mole of acid would titrate with 2 mole base.

We know this because of the 2 in front of the NaOH in other words, if you had spilled one mole of H2SO4 on your skin, you would need 2 mole of NaOH to stop the burning ... (Don't nobody try this).

So just because you have strong acid + strong base, you can not assume a 1:1 ratio for titration. Professors like to trip you up that way

When balancing an equation, don't freak out if you can't remember the charge on a polyatomic ion. Say you have the poly (SO4) Hmm, the charge is? Well here is a hint: on the test they give you the starting as in example above H2SO4 + 2NaOH---> ???

From the above we know SO4 has a charge of 2 by using the criss-cross method for chemical formulas. (OH) has to have a charge of -1 since we know Na is +1 and formula is written as NaOH.

You may know this stuff already, but thought I would throw it out there anyways.

**final**might not be a 1:1 molar ratio as it was for HCL. For example it could have been H2SO4 in which case you would have:H2SO4 + 2NaOH ----> Na2SO4 + 2H2O

From the equation above you can see that 1 mole of acid would titrate with 2 mole base.

We know this because of the 2 in front of the NaOH in other words, if you had spilled one mole of H2SO4 on your skin, you would need 2 mole of NaOH to stop the burning ... (Don't nobody try this).

So just because you have strong acid + strong base, you can not assume a 1:1 ratio for titration. Professors like to trip you up that way

When balancing an equation, don't freak out if you can't remember the charge on a polyatomic ion. Say you have the poly (SO4) Hmm, the charge is? Well here is a hint: on the test they give you the starting as in example above H2SO4 + 2NaOH---> ???

From the above we know SO4 has a charge of 2 by using the criss-cross method for chemical formulas. (OH) has to have a charge of -1 since we know Na is +1 and formula is written as NaOH.

You may know this stuff already, but thought I would throw it out there anyways.

- alextemplet
- King Cobra
**Posts:**5599**Joined:**Fri Dec 23, 2005 4:50 pm**Location:**South Louisiana (aka Cajun Country)-
**Contact:**

Another good trick for polyatomic ions is that, with a few exceptions like OH-, they almost always have the same charge as the normal non-oxygen ion. For example, all polys with chlorine (ClO-, ClO2-, ClO3-, ClO4-) all have a charge of -1, just like the chloride ion (Cl-). All polys of sulfur (SO3-2, SO4-2) have the same -2 charge as the sulfide ion (S-2).

Generally speaking, the more people talk about "being saved," the further away they actually are from true salvation.

~Alex

#2 Total Post Count

~Alex

#2 Total Post Count

### Who is online

Users browsing this forum: No registered users and 2 guests