how do do this chem q (2 Viewers)

[jonhysmitw2]

 

[Grand Master IUPAC]

Since it's strong acid strong base, it will completely ionise with a final pH of about 7 (usually). So you need some excess hydrochloric acid which will lower pH down to 1.

Basically, figure out the volume of HCl needed to fully neutralise the NaOH, then using moles and stuff, figure out how much volume of excess HCl you need for a pH of 1. Then add the volumes together. I think that's how it works.

 

[SadCeliac]

View attachment 41117

pH of 1 means [H+] of 10^-1 = 0.1M
Hence you need a final number of HCL moles in XS to produce that 0.1M concentration

When v added is 30mL you have a total v of 55mL, and you're adding 0.055mol HCl against 0.025mol NaOH
Hence 0.03mol H+ in excess and then solve accordingly

 

[jonhysmitw2]

pH of 1 means [H+] of 10^-1 = 0.1M
Hence you need a final number of HCL moles in XS to produce that 0.1M concentration

When v added is 30mL you have a total v of 55mL, and you're adding 0.055mol HCl against 0.025mol NaOH
Hence 0.03mol H+ in excess and then solve accordingly

how do yk its 30ml added tho (not looking at answer)

 

[SadCeliac]

how do yk its 30ml added tho (not looking at answer)

Probably try each one out honestly
It's a multi

 

[synthesisFR]

View attachment 41117

So the general stuff: HCL is monoprotic and has the same concentration as the base in this case right. So you need 25 ml of HCl to make a neutral solution.
Now you want the final concentration to be 0.1M and the volume the current neutral solution is 50ml or 0.05 L
so u find the moles of hcl needed to get this concentration: n = c.v = 0.1x0.05 = 0.005 mol
then find the volume of hcl needed to produce this many moles:
v = n/c = 0.005/1 = 0.005L = 5ml
so u add to get 30 ml in total
(Someone confirm but idk why this method works like to me it doesn't make sense bc ur adding aqueous hcl so i dont know how my first n = vc thing works to get the moles required to make the concentration within 50ml its a bit confusing @carrotsss help)
I would just check between b c d after the first step since its mcq ....

 

[STBAccuracy]

I did a bit of maths for this Q...

 

[synthesisFR]

ok yes u got the correct method
idk why mine works even tho its technically flawed (@carrotsss help i alr tagged u buffoon)
anyways yea its better to just use mcq answers

 

[carrotsss]

So the general stuff: HCL is monoprotic and has the same concentration as the base in this case right. So you need 25 ml of HCl to make a neutral solution.
Now you want the final concentration to be 0.1M and the volume the current neutral solution is 50ml or 0.05 L
so u find the moles of hcl needed to get this concentration: n = c.v = 0.1x0.05 = 0.005 mol
then find the volume of hcl needed to produce this many moles:
v = n/c = 0.005/1 = 0.005L = 5ml
so u add to get 30 ml in total
(Someone confirm but idk why this method works like to me it doesn't make sense bc ur adding aqueous hcl so i dont know how my first n = vc thing works to get the moles required to make the concentration within 50ml its a bit confusing @carrotsss help)
I would just check between b c d after the first step since its mcq ....

the amount the concentration will be reduced beyond it is pretty negligible, you only need about 0.5mL more to account for it as @STBAccuracy shower. Imo the answer should really be D though since C doesn’t actually reach the pH required, D is the only one which exceeds it