GC question

[UnimaasMED]

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If 20 ml of 0.012 M solution of Ca(OH)2 is added to 48 ml of HBr, what is the concentration of the HBr?


I used the N1V1=N2V2 formula but used 68ml for V2. The correct answer had 48ml for V2. I am confused. In which case do we add up the volume and in which case we don't?

 

[CANgnome]

I think the type of calculation question where you add up the volume is the one where you titrate X with Y and then establish a new pH?

In this case, it is just a simple 1:1 so you do not have to add things up.

 

[nyr201]

If 20 ml of 0.012 M solution of Ca(OH)2 is added to 48 ml of HBr, what is the concentration of the HBr?


I used the N1V1=N2V2 formula but used 68ml for V2. The correct answer had 48ml for V2. I am confused. In which case do we add up the volume and in which case we don't?

usually when it asks what volume is required to give a certain conc so when u basically have M1 M2 and a V so then ud figure out the other V and subtract it from the first V and that'd be the V required to give u ur desired M...to clarify for example if u have 100 mL of .10 M Juice (lame ex sorry lol) and u want to make it .05M how much V of (water in this case since its a dilution) so u have (100 mL)x(.10M)= (.05M)x (V2) answer is 200 mL so you subtract 200-100 and what u get is 100 mL of water to get the conc down to .05 M...so this is when u mess around with the volumes...in ur case u u have two volumes so u leave them as is and just plug and chug to get ur M2...hope this helps 😀

 

[UnimaasMED]

usually when it asks what volume is required to give a certain conc so when u basically have M1 M2 and a V so then ud figure out the other V and subtract it from the first V and that'd be the V required to give u ur desired M...to clarify for example if u have 100 mL of .10 M Juice (lame ex sorry lol) and u want to make it .05M how much V of (water in this case since its a dilution) so u have (100 mL)x(.10M)= (.05M)x (V2) answer is 200 mL so you subtract 200-100 and what u get is 100 mL of water to get the conc down to .05 M...so this is when u mess around with the volumes...in ur case u u have two volumes so u leave them as is and just plug and chug to get ur M2...hope this helps 😀

I am still confused. It doesn't make sense. When we are solving for the concentration of a solute upon mixture we have to plug the final volume right?

 

[nyr201]

I am still confused. It doesn't make sense. When we are solving for the concentration of a solute upon mixture we have to plug the final volume right?

ok for ur example u multiply .012 by 2 to get normality so its .024 N

so now the M1V1=M2V2 plug and chug.....

(.024N) x (20 mL) = (M2) x (48 mL)

= (.024) x (20 mL)/ 48 mL = M2 ur answer is .01 M or N doesnt matter since M=N in this case bc HBr is monoprotic...

ok so the reason u dont add the volumes is bc 48 mL is already ur final volume already its at that volume that HBr is .01 M...hope this clarifies things 😀

 

[quillow]

ok for ur example u multiply .012 by 2 to get normality so its .024 N

so now the M1V1=M2V2 plug and chug.....

(.024N) x (20 mL) = (M2) x (48 mL)

= (.024) x (20 mL)/ 48 mL = M2 ur answer is .01 M or N doesnt matter since M=N in this case bc HBr is monoprotic...

ok so the reason u dont add the volumes is bc 48 mL is already ur final volume already its at that volume that HBr is .01 M...hope this clarifies things 😀

The question asked if 20 ml of Ca(OH)2 is added to 48 ml HBr... so the final volume would be 68 ml, wouldn't it? I don't think you have to consider final volume in this question though.

I answered this question by figuring they were (indirectly) asking about a titration (what is the concentration of HBr needed to titrate the solution of calcium hydroxide?). So I calculated moles of OH-, which should equal moles of H+ (at equivalence point). Essentially the same calculation as above to find concentration of HBr, but that's why I didn't consider final (combined) volume. Please correct me if I'm wrong though.