Titration Question

[NRAI2001]

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How do u figure out the pH of an acid that has been titrated by a base?

For example:

Your titrating a .100M sample of HCO2OH with .100M NaOH; you have 50mL of acid;

what is pH after 60 ml of NaOH?

Do u use the neutralization equation?

a x [A] x Va = b x x Vb

This is how i was thinking of doing it, but I can't seem to get the right answer. Calculate how much H is left after titration is complete. then take the -log of the concentration.

pH = -log((bVb)/([A]Va))

 

[theblastopore]

How do u figure out the pH of an acid that has been titrated by a base?

For example:

Your titrating a .100M sample of HCO2OH with .100M NaOH; you have 50mL of acid;

what is pH after 60 ml of NaOH?

Do u use the neutralization equation?

a x [A] x Va = b x x Vb

This is how i was thinking of doing it, but I can't seem to get the right answer. Calculate how much H is left after titration is complete. then take the -log of the concentration.

pH = -log((bVb)/([A]Va))

Hi,
If it's strong acid/strong base reaction then you can do some thing like that because we assume the reaction will go to completion.

In this case, I think you have an equilibrium. Review your acid-base equilibrium (toward the end of chem 2)

I'm sure some one else will give you a better answer soon. If not, I'll to explain it some time later (gotta review it first )

 

[ShyRem]

Personally I always go back to mass balance/charge balance and simultaneous equations. Always made more sense to me this way.

 

[XT777]

All of the HA is neutralized, as is the first 50 ml of NaOH (both .1M).
Therefore the solutiuon would be basic because of the 10 additional ml of NaOH not neutralized.

Calculate moles of OH- in 10 ml of .1M NaOH (.01L*.1M). <--presumed completely ionized.

divide the moles of OH- by .110L (<--total volume of solution -- .05L+.06L) to get molarity of OH-

14 - (-log[OH-]) = answer in pH (~12)


is this an MCAT type Q<--to be done without a calculator?

what are the answer choices? <---is only one of the answers a basic pH?

How do u figure out the pH of an acid that has been titrated by a base?

For example:

Your titrating a .100M sample of HCO2OH with .100M NaOH; you have 50mL of acid;

what is pH after 60 ml of NaOH?

Do u use the neutralization equation?

a x [A] x Va = b x x Vb

This is how i was thinking of doing it, but I can't seem to get the right answer. Calculate how much H is left after titration is complete. then take the -log of the concentration.

pH = -log((bVb)/([A]Va))

 

[XT777]

actually...I think the answer i just gave is wrong since carbonic acid is diprotic.

They didn't give any Ka values?

Which questionare you trying to answer?

"How do u figure out the pH of an acid that has been titrated by a base?"

or

"what is pH after 60 ml of NaOH?"


For the second:
All of the 50 ml of HA are neutralized, as is the first 50 ml of NaOH (both .1M).
Therefore the solutiuon would be basic because of the 10 additional ml of NaOH not neutralized.

Calculate moles of OH- in 10 ml of .1M NaOH (.01L*.1M). <--presumed completely ionized.

divide the moles of OH- by .110L (<--total volume of solution -- .05L+.06L) to get molarity of OH-

14 - (-log[OH-]) = answer in pH (~12)


is this an MCAT type Q<--to be done without a calculator?

what are the answer choices? <---is only one of the answers a basic pH?

 

[NRAI2001]

All of the HA is neutralized, as is the first 50 ml of NaOH (both .1M).
Therefore the solutiuon would be basic because of the 10 additional ml of NaOH not neutralized.

Calculate moles of OH- in 10 ml of .1M NaOH (.01L*.1M). <--presumed completely ionized.

divide the moles of OH- by .110L (<--total volume of solution -- .05L+.06L) to get molarity of OH-

14 - (-log[OH-]) = answer in pH (~12)


is this an MCAT type Q<--to be done without a calculator?

what are the answer choices? <---is only one of the answers a basic pH?

This Q wasn't taken directly from anything directly related to the MCAT. My cousin just took his chem final and he asked me if i knew how to do this. But I think it is a really good question now that I ve been working on it.

I came to the same answer that u did. I assumed carbonic acid to be monoprotic, i don't know if that is a mistake? But I guess the only way to know if carbonic acid would be have in a monoprotic fashion would be to compare the pka between sodium hydroxide and the 2nd acid site of the carbonic acid. I don't have the exact values with me, and I don't feel like looking it up, but I m guessing that the 2nd acid site is probably less acidic than the water that the sodium hydroxide would form. So it would behave in a monoprotic fashion.

Please correct me if I am wrong. 😀