Chemistry Questions
Started by klpennin
Did you just erase your answer for the first question???
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yeah I erased it, I realised it was wrong....
I assumed the final product was in 1L because i said 0.02 M. But really its not in 1L, not even close. I don't know how to factor in the variable Total litres the solution will be in... I was thinking a quadratic but it didn't work out...
IF the answer is 2000 mL then here is how to do it:
(20mL -OH)(1M -OH)=V of HCl (1M HCl)
solve and volume of HCl needed to titrate -OH completely is 20mL.
this makes sense because at equilibrium, the mole (and molarity) of the acid and base are equal.
Now, we want to make the solution acidic, at a pH of 2..meaning we want the solution to have a H+ concentration of 1x10^-2. So, we just use c1v1=c2v2 again.
(20mL HCl)(1M HCl)=V of HCl (1x10^-2 M HCl)
V=2x10^3
let me know if that is correct!
(20mL -OH)(1M -OH)=V of HCl (1M HCl)
solve and volume of HCl needed to titrate -OH completely is 20mL.
this makes sense because at equilibrium, the mole (and molarity) of the acid and base are equal.
Now, we want to make the solution acidic, at a pH of 2..meaning we want the solution to have a H+ concentration of 1x10^-2. So, we just use c1v1=c2v2 again.
(20mL HCl)(1M HCl)=V of HCl (1x10^-2 M HCl)
V=2x10^3
let me know if that is correct!
I got 20.04ml, am I right?
I think what we have to see first is that when 20 ml of NaOH and 20 ml of HCl is mixed together, the ph will be 7. but question asks for the vol. of HCl when pH is 2. so i think we have to do (40ml)(1x10^-7)=V(1x10^-2)
and find V. and add 20ml to that.
no????
and find V. and add 20ml to that.
no????
I approched it in a different way but i think ur way is correct too
20ml + 20ml of same molarity = 40ml of neutral
since we want it to be acidic, we need to add more of acid than base.
1M (X) = 0.01M (40+X)
X = 0.04 + 0.01X
0.99X = 0.04
X = 0.0404
40 + 0.0404 - 20 = 20.0404
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i found this from other thread..
i dont know how to quote post from other thread so i am just copying and pasting it......
its from osimsDDS
this question is infamous...
pH of 2 = 1x10^-2
Neutralization between the HCl and the NaOH will give you 20ml, add that to 20ml you started out with and you get 40 ml...
Now you use this formula (many people do it differently but i find this the best way for me):
(HCl M)(x) = (pH)(total volume + x)
(1M)(x) = (1x10^-2)(40 + x)
1x= .4 + .01x
x= .4ml
Now add that to the starting volume of HCl which was 20 ml so your answer is 20.4ml
i dont know how to quote post from other thread so i am just copying and pasting it......
its from osimsDDS
this question is infamous...
pH of 2 = 1x10^-2
Neutralization between the HCl and the NaOH will give you 20ml, add that to 20ml you started out with and you get 40 ml...
Now you use this formula (many people do it differently but i find this the best way for me):
(HCl M)(x) = (pH)(total volume + x)
(1M)(x) = (1x10^-2)(40 + x)
1x= .4 + .01x
x= .4ml
Now add that to the starting volume of HCl which was 20 ml so your answer is 20.4ml
I am working on it. Give me about 20 min...
I know that my way was the correct way of solving this problem but I know what you are thinking and not sure why it gives different answer.
I will figure it out...give me a sec 🙂
NH3's molecular formula ie trigonal pryamidal because of the lone pair which creates an angle less than 120.
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