[candidateof]

Hi everybody,

I am facing a problem in solving this question, help if you know please.

What mass of Al(NO3)3 has to be added to a 0.250 m aqueous solution of NaCl that contains 850 g of solvent in order for the solution to have an ionic strength of 0.420 m?
(Atomic masses: Al:26.98, N: 14, O:16)

The choices are:

a) 6.03 g
b) 4.53 g
c) 4.53 kg
d) 5.13 g
e) 5.13 kg

----------------
I have tried to answer this question and I am not getting one of those answers! What I tried:
I: Ionic strength= 0.5*sum(bz)

I(solution)=I(Al(NO3)3) + I(NaCl)
I(NaCl)= 0.5*0.25=0.125 m (since z+=|z-|=1)
I(solution)=0.42 m
=> I(Al(NO3)3)=0.42-0.125=0.295 m = 0.295 moles of Al(NO3)3/Kg solvent
=> multiplying by 0.85 kg solvent, we need 0.25075 moles of Al(NO3)3
From MW=> 53.159 g

and it is not there! I am missing something but I didn't figure it out yet.

Thanks in advance

[Jasim]

"I(Al(NO3)3)=0.42-0.125=0.295 m = 0.295 moles of Al(NO3)3/Kg solvent"

You are saying Ionic Strength of Al(NO3)3 = 0.295, which looks correct to me.

I think you are missing a step. Plugging this into the equation for ionic strength of Al(NO3)3:
0.295  = (0.5) * Molality of Al(NO3)3

EDIT: wait, I was using your numbers. How about this: I: Ionic strength= 0.5*sum(bz²)
And what is the z for Al(NO₃)³

Also, disclaimer, I'm trying to learn this too, so don't take what I post for final answer.

[AWK]

For NaCl I=c
For Al(NO₃)₃ I=c/6



addition
sorry my fault - should be 6c

[candidateof]

Hi Jasim,

I believe that I have tried that too, but let us see. for Al(NO3)3: z+=3 and z-=-1
For Al(NO3)3:
I=0.5*sum(b z'sqr')
I=0.5*(9+1)* molality of Al(NO3)3      "I=0.295"
=>molality of Al(CO3)3=0.059      "0.85 kg solvent"
=>0.05015 moles of Al(CO3)3     "MW=212"
=>10.6318 g of Al(CO3)3

What do you think? Please let me know if you have a new idea.

[candidateof]

Hello AWK,

How come?

For NaCl cause z+=|z-|=1
so, I=0.5*c

but for Al(CO3)3, z+=3 and |z-|=1
so, I=0.5*c*10
10 came from: (3 sqr + 1 sqr)= 9+1

[AWK]

Hello AWK,

How come?

For NaCl cause z+=|z-|=1
so, I=0.5*c

but for Al(CO3)3, z+=3 and |z-|=1
so, I=0.5*c*10
10 came from: (3 sqr + 1 sqr)= 9+1

NaCl consists of two ions with charge 1+ and 1-
Al(NO3)3 consists of one ion 3+ and 3 ions 1-
I=1/2 Sum od c_iZ_i²

[candidateof]

Hello AWK,

But I only use z+ and z- in the ionic strength formula, right? I don't use number of ions?

[AWK]

for NaCl
I = 1/2(c x 1² + c x 1² ) = c
for Al(NO₃)₃
I = 1/2(3 x c x 1² + c x 3² ) = 6c

[candidateof]

for NaCl
I = 1/2(c x 1² + c x 1² ) = c
for Al(NO₃)₃
I = 1/2(3 x c x 1² + c x 3² ) = 6c

Thank you very much sir. I got the answer now.

according to your formulas:

I(NaCl)=0.25m
I(Al(NO3)3)=6b
also I(Al(NO3)3)=0.42-0.25=0.17m
=> b=0.02833  "0.85 kg solvent"
=> 0.0241 moles of Al(NO3)3  "MW=212.996506"
=> 5.13 g, which is d)

Thank you Jasim as well

[AWK]

Look at the attached table

[candidateof]

Look at the attached table

I really appreciate your help. The file is super helpful

Regards,