G-chem Molarity from Titration?

[Dencology]

11.6 mL of 3.0 M sulfuric acid are required to neutralize the sodium hydroxide in 25mL of NaOH. what is the molarity of the NaOH?

Ans. 2.8 M NaOH

so i was thinking to use NaVa= NbVb

so Nb= (11.6)(6)/25 which is 2.8=Nb. which is the same as Molarity.

---

Members don't see this ad.

 

[zuma35]

Yes.

 

[chessxwizard]

For a question like this where the second acidic proton's contribution to the pH is assumed to be negligible, you just do M1V1=M2V2. This is not the same for bases such as Ca(OH)2, since those totally dissociate in aqueous medium. In that case, you would use the formula with normality substituted for molarity.

 

[Dencology]

For a question like this where the second acidic proton's contribution to the pH is assumed to be negligible, you just do M1V1=M2V2. This is not the same for bases such as Ca(OH)2, since those totally dissociate in aqueous medium. In that case, you would use the formula with normality substituted for molarity.

if you do M1V1=M2V2 then you get something like 1.4 M which is not the correct answer. I guess from writing the reaction we see that we have 2 moles of NaOH so we need to multiply 1.4 with 2 to get 2.8?

what do you think?

 

[zuma35]

For titrations you always use N1V1=N2V2.

 

[Dencology]

For titrations you always use N1V1=N2V2.

so i guess i was right at the beginning. right?

 

[zuma35]

Yup

 

[Sublimation]

I havent read any of the study material yet, so im going off what i recall from Gchem, but i just used Volume of HCl X Molarity=Moles

to Neutralize the base Moles of Acid = Moles of Base right?

You have Moles of base divide by the .025L remember that the molarity is always in moles/liter. Now do u divide by .025L or .0116+.025? im confused as to whether or not we add the acid?

 

[Sublimation]

I havent read any of the study material yet, so im going off what i recall from Gchem, but i just used Volume of HCl X Molarity=Moles

to Neutralize the base Moles of Acid = Moles of Base right?

You have Moles of base divide by the .025L remember that the molarity is always in moles/liter. Now do u divide by .025L or .0116+.025? im confused as to whether or not we add the acid?

HAHA lol man my bad, lol i didnt realize it said sulfuric acid, lol i was using my method and when i multiplied it by 2 i got the right answer so i read it again lol, i think i should just go to bed 2AM here on the east coast lol.

 

[Dencology]

For titrations you always use N1V1=N2V2.

Zuma35,
so when we do NaVa=NbVb, we get normality. Now how is it that it is equal to the molarity?

 

[zuma35]

Normality is the molarity multiplied by the number of acid or base equivalents. So for a monoprotic acid or base normality equals molarity. For diprotic normality is double molarity, etc.

 

[Dencology]

Normality is the molarity multiplied by the number of acid or base equivalents. So for a monoprotic acid or base normality equals molarity. For diprotic normality is double molarity, etc.

So for a rxn like this:

2 NaOH + H2SO4----> Na2SO4 + 2 H2O
.
so since we have 2 NaOH and we find the normality to be 1.4 then we multiply it by two because we have 2 moles of NaOH and thats how we get 2.8, right?

 

[zuma35]

No. Since H2SO4 has two H's the Normality is 2 times the molarity (2 * 3M) = 6N. You did it right in the first post. You get 2.8 for the molarity of NaOH, which is also the N because NaOH only gives off 1 equivalence of base.

 

[Dencology]

No. Since H2SO4 has two H's the Normality is 2 times the molarity (2 * 3M) = 6N. You did it right in the first post. You get 2.8 for the molarity of NaOH, which is also the N because NaOH only gives off 1 equivalence of base.

Ok. i got you now.