Gen Chem Acid/Base Ques

[deleted92121]

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I feel really stupid on this one. Not looking for an answer, just a way to set it up so I can solve it. I feel like I am missing some basic step that is messing up my process.

How much 1.20 M NaOH is needed to neutralize 225 mL of 3.0 M H2SO4?

 

[Erek94]

I think with sulfuric acid although it is diprotic acid the first hydrogen acts as a strong acid and thus making it dissociate the first hydrogen to the full extent, I'm not sure but probably not the same case with that second hydrogen. So I'm assuming you can use the m1v1= m2v2 equation and remember to keep the volume in ml or convert to liters.

 

[UrshumMurshum]

The second hydrogen of sulfuric acid has a pka of 1.99 so it's considered a weak acid.

The way I think about it is you have ~3 times as much concentration of sulfuric acid as you do NaOH. Therefore you're going to need about ~3 times as much NaOH to neutralize it. 225 *3 = 775.

The answer is going to be less than 775 mL since our approximation didn't include the .2 in the 1.2 of NaOH.

Not sure if this is right but that's my initial reaction to this problem.

 

[howlovely]

Wow, I've forgotten a lot of chemistry 🙁

I'd start with balancing the reaction equation: 2NaOH + H2SO4 -> Na2SO4 + 2H2O. From there, you can figure out the ratio of Na+ and (SO4)2- you need, which is 2:1. Then convert both to moles.

 

[gettheleadout]

Although the second proton in H2SO4 is weakly acidic, in the presence of a strong base like NaOH, both protons will be stripped from the acid. Thus, neutralization of H2SO4 requires a 2:1 mole ratio of strong base to acid.

Full answer and explanation in white below:

Beginning with 0.225 L of 3.0 M H2SO4, we have between 1/5 and 1/4 L containing 3 moles of acid, meaning we have between .6 and .75 moles H2SO4 to be neutralized. With a 2:1 mole ratio of base to acid, we need between 1.2 and 1.5 moles of NaOH. With a stock solution of 1.2 M NaOH, 1 L of stock would give us the lower end of the range of the needed value (1.2 moles NaOH) and the upper end of the range (1.5 moles NaOH = 1.2 + 1.2/4) would be 1.25 L.

Since 0.225 L (our starting volume of acid) is in fact the exact midpoint between the range we established, we can conclude that our needed volume of NaOH is precisely between 1 and 1.25 L.

1.125 L of 1.2 M NaOH are needed to neutralize 0.225 L of 3.0 M H2SO4.

 

[rjosh33]

Although the second proton in H2SO4 is weakly acidic, in the presence of a strong base like NaOH, both protons will be stripped from the acid. Thus, neutralization of H2SO4 requires a 2:1 mole ratio of strong base to acid.

Full answer and explanation in white below:

Beginning with 0.225 L of 3.0 M H2SO4, we have between 1/5 and 1/4 L containing 3 moles of acid, meaning we have between .6 and .75 moles H2SO4 to be neutralized. With a 2:1 mole ratio of base to acid, we need between 1.2 and 1.5 moles of NaOH. With a stock solution of 1.2 M NaOH, 1 L of stock would give us the lower end of the range of the needed value (1.2 moles NaOH) and the upper end of the range (1.5 moles NaOH = 1.2 + 1.2/4) would be 1.25 L.

Since 0.225 L (our starting volume of acid) is in fact the exact midpoint between the range we established, we can conclude that our needed volume of NaOH is precisely between 1 and 1.25 L.

1.125 L of 1.2 M NaOH are needed to neutralize 0.225 L of 3.0 M H2SO4.

👍

OP, in acid-base titrations, it's almost always helpful to find the equivalence volume (amount of titrant needed to neutralize analyte) first. Since sulfuric acid is diprotic, there will be two equivalence volumes (one for the first proton of H2SO4, another for the proton of HSO4-). Using C1V1=C2V2 and some algebra (with C being concentration and V being volume), we get V1 = (3.0 M)(225 mL)/(1.2 M) = 562.5 mL for the first equivalence volume.

Now comes the important part. The second equivalence volume is always going to be twice the first equivalence volume. So, the volume of NaOH at the second equivalence point is (2 x 562.5 mL) = 1,125 mL, or 1.125 L. This is the volume of NaOH needed to completely neutralize H2SO4. Note, it is the same as getthe stated above (exactly between 1 L and 1.25 L).

 

[johnnyscans]

I feel really stupid on this one. Not looking for an answer, just a way to set it up so I can solve it. I feel like I am missing some basic step that is messing up my process.

How much 1.20 M NaOH is needed to neutralize 225 mL of 3.0 M H2SO4?

H2SO4 is treated like a diprotic acid. For the first neutralization
(1.2)(x) = (.225)(3)
x is amount of mL to reach the first equivalence point. .5x is the half-equivalence point (which may be important in other questions.)
2x is the amount to fully neutralize the acid. The is the second equivalence point, and it is always 2x the first equivalence point.

 

[rjosh33]

Is there an echo in here? 🙂

 

[johnnyscans]

Is there an echo in here? 🙂

Echo Echo Echo

 

[rjosh33]

👍