BR chem ch 5 page 323 number 29

[2010premed]

25 mL of an unknown acid when titrated by exactly 30 mL of 0.1 M KOH, required seven drops of .1 M HCl to return to equivalence. What is the concentration of the unknown acid?
A 0.1217 M acid
B. 0.1183 M acid

I don't really understand this questions... I did .1M (30mL) / 25mL = .12 M...

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[loveoforganic]

Yeah, you get .12, but you needed a few drops of acid to return to equivalence (you overshot equivalence point). Therefore, your acid is giong to be slightly less concentrated than what you calculate using the amount of KOH you added (answer is B).

 

[2010premed]

I guess I'm not seeing why it would be less concentrated, could you please clarify this point?

 

[loveoforganic]

You "accidently" used too much KOH initially. They imply this by saying you had to add a few drops of HCl to return to equivalence. You do your calculations figuring in more KOH than you actually needed, so your true concentration is slightly weaker than what you come up with.

 

[MrNeuro]

You "accidently" used too much KOH initially. They imply this by saying you had to add a few drops of HCl to return to equivalence. You do your calculations figuring in more KOH than you actually needed, so your true concentration is slightly weaker than what you come up with.

i thought of it in the sense that the addition of 7 drops of .1 M HCl decreases the concentration because .12 M soln + .1 M HCl brings down the concentration is this totally wrong?

 

[MedPR]

25 mL of an unknown acid when titrated by exactly 30 mL of 0.1 M KOH, required seven drops of .1 M HCl to return to equivalence. What is the concentration of the unknown acid?
A 0.1217 M acid
B. 0.1183 M acid

I don't really understand this questions... I did .1M (30mL) / 25mL = .12 M...

Yea, but you aren't accounting for the HCl added to bring it to equivalence.

If the concentration of the unknown acid was .12M, the 30mL of KOH would've been perfect to bring it to equivalence, but since you had to add more acid after adding base, you know that 30mL of KOH was a little too much. That means that the unknown acid must have a concentration a little less than what you expected (.12M). So B.

If the unknown acid was 0.1217M, then 30mL of KOH wouldn't have been enough.

 

[MedPR]

i thought of it in the sense that the addition of 7 drops of .1 M HCl decreases the concentration because .12 M soln + .1 M HCl brings down the concentration is this totally wrong?

Adding HCl decreases the concentration of the KOH and HA in the final solution since it adds to the total volume. However, it doesn't affect the original concentration of HA. At least I don't think it does..

 

[loveoforganic]

Adding HCl decreases the concentration of the KOH and HA in the final solution since it adds to the total volume. However, it doesn't affect the original concentration of HA. At least I don't think it does..

correct

 

[MrNeuro]

Adding HCl decreases the concentration of the KOH and HA in the final solution since it adds to the total volume. However, it doesn't affect the original concentration of HA. At least I don't think it does..

correct

so just to clarify you could technically solve for the volume contained in 7 drops of HCl given that the answer is B

 

[MedPR]

so just to clarify you could technically solve for the volume contained in 7 drops of HCl given that the answer is B

Yup. The key to the HCl mentioned in the question is that you added some volume (7 drops) of acid to reach equilibrium. That additional acid tells you that the concentration of HA is less than 0.12 calculated from the volume of KOH used.

It's a calculation you probably won't have to make on the MCAT, but yes you could calculate it.