AAMC 4 titration

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chiddler

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The NaOH (aq) was standardized by each student by titration against a pure sample of KHP, as shown in the equation...

I don't understand what this is saying. What does it mean to standardize a solution by titration?

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I think it just means they developed a titration curve from the NaOH that was given against the KHP. Standardized in the sense that during a titration you use a known concentration of titrant of which to compare against an unknown.
 
"developed a titration curve from the NaOH that was given against the KHP"?

can you please explain?
 
In a nutshell, a standardized titration means that they plotted a graph which shows the pH of different combinations of NaOH and KHP. The equation should tell you what the pH of the solution is for any combination of NaOH and KHP.
 
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In a nutshell, a standardized titration means that they plotted a graph which shows the pH of different combinations of NaOH and KHP. The equation should tell you what the pH of the solution is for any combination of NaOH and KHP.

i know what a titration curve is :-x

i'm trying to understand the meaning of the sentence i wrote.
 
Oh. Isn't it just the equation of the curve then?

I'm not sure. Specifically, what does "that was given against KHP" mean?

If you can offer a different way of explaining the original sentence I wrote, would be lovely.
 
I'm not sure. Specifically, what does "that was given against KHP" mean?

If you can offer a different way of explaining the original sentence I wrote, would be lovely.

It should be an equation of the pH of a sample of NaOH vs the amount of pure KHP added into the solution. pH as a function of KHP added to the solution.
 
To understand the original sentence you need to understand that an aqueous solution of NaOH does not have a stable pH over time. The solution gradually acidifies due to the CO2 in air. Therefore the pH must be determined precisely prior to use. For this, we use KHP titration as the traditional method. KHP is a solid, easily handled, and stable to air and therefore is useful for this purpose.

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In other words "standardize by titration" means to determine the pH of the standard NaOH solution.

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Funny, I just got done with this practice test about 10 minutes ago, and did a double take at this sentence too.

I'm not sure. Specifically, what does "that was given against KHP" mean?

If you can offer a different way of explaining the original sentence I wrote, would be lovely.

Not sure if you've gotten your answer or not, but the text from the passage is basically saying "the molarity of the NaOH was determined using KHP in a titration." This fact is then used to answer question 26.
 
It should be an equation of the pH of a sample of NaOH vs the amount of pure KHP added into the solution. pH as a function of KHP added to the solution.

i don't think that's true because NaOH concentration was given to me. furthermore, KHP is a solid and was made into solution prior to adding.

furthermore, I think you're misunderstanding the "equation" part of the statement. the equation that was given is a chemical equation, not a mathematical one.

like CH4 + O2 ==> CO2 + H2O equation. not a x=y equation.

To understand the original sentence you need to understand that an aqueous solution of NaOH does not have a stable pH over time. The solution gradually acidifies due to the CO2 in air. Therefore the pH must be determined precisely prior to use. For this, we use KHP titration as the traditional method. KHP is a solid, easily handled, and stable to air and therefore is useful for this purpose.

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Ok! I like this. Is this possible when KHP concentration is unknown?
 
Funny, I just got done with this practice test about 10 minutes ago, and did a double take at this sentence too.



Not sure if you've gotten your answer or not, but the text from the passage is basically saying "the molarity of the NaOH was determined using KHP in a titration." This fact is then used to answer question 26.

Are you sure? Because we are given 8g is dissolved in 2 L. My understanding is that the acid is unknown.
 
Are you sure? Because we are given 8g is dissolved in 2 L. My understanding is that the acid is unknown.

Yeah, for reasons that gaystains mentioned the original molarity of the NaOH can fluctuate with time. To determine it precisely, it has to be titrated into a known amount (mass) of KHP (or some other standard) to determine its exact molarity.

This determined molarity is then used to determine the moles present (and subsequently, the molecular weight) of the unknown acid in a separate titration.
 
Yeah, for reasons that gaystains mentioned the original molarity of the NaOH can fluctuate with time. To determine it precisely, it has to be titrated into a known amount (mass) of KHP (or some other standard) to determine its exact molarity.

This determined molarity is then used to determine the mass of the unknown acid in a separate titration.

OH! OH DAMN!

I see!

thanks!

I'd appreciate one more bit of help, please. The question within the image I just posted.
 
OH! OH DAMN!

I see!

thanks!

I'd appreciate one more bit of help, please. The question within the image I just posted.

This is tricky to explain, so bear with me.

If there's some moisture present in the KHP it's going to read out as heavier than the dry KHP actually is when you weigh it on the scale prior to the titration. The problem with this is that the mass of KHP you record is then converted into moles of KHP. Because the moles of KHP is used to get the moles of NaOH (in a one-to-one stoichiometry) in the titration, then the moles of NaOH you get from the titration will be calculated as being higher than there is actually present. This leads to a determined molarity that is actually higher than it should be.

Does that make sense or did I make it worse?
 
This is tricky to explain, so bear with me.

If there's some moisture present in the KHP it's going to read out as heavier than the dry KHP actually is when you weigh it on the scale prior to the titration. The problem with this is that the mass of KHP you record is then converted into moles of KHP. Because the moles of KHP is used to get the moles of NaOH (in a one-to-one stoichiometry) in the titration, then the moles of NaOH you get from the titration will be calculated as being higher than there is actually present. This leads to a determined molarity that is actually higher than it should be.

Does that make sense or did I make it worse?

Too much acid is calculated due to extra water weight, so the NaOH is assumed to be too concentrated. Aha.

Thank you for the very helpful explanation.
 
This is tricky to explain, so bear with me.

If there's some moisture present in the KHP it's going to read out as heavier than the dry KHP actually is when you weigh it on the scale prior to the titration. The problem with this is that the mass of KHP you record is then converted into moles of KHP. Because the moles of KHP is used to get the moles of NaOH (in a one-to-one stoichiometry) in the titration, then the moles of NaOH you get from the titration will be calculated as being higher than there is actually present. This leads to a determined molarity that is actually higher than it should be.

Does that make sense or did I make it worse?

Awesome explanation.
 
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