Concentration question VERY HARD (1 Viewer)

[coolcats]

Hey guys,
Could you please help with this q?
23.0 ml of 1.0 M potassium iodide is mixed with 77.0 ml of 1.0 M lead (II) nitrate. A brightly coloured precipitate of lead iodide is observed to form. Calculate the concentration of excess lead ions remaining in solution:

SDA

 

[Sachin123]

I am struggling with this question is this from James ruse???
Much help needed.
TY

 

[lpodtouch]

Have they provided you with an answer? I'm not sure whether my answer is correct ;P

 

[Sachin123]

Can you post your answer and working Ipodtouch

 

[HeroicPandas]

I don't want to give you too much... but I'd like to send you in the right direction

Firstly, write down the balanced chemical equation:

2KI (aq) +Pb(NO3)2 (aq) -> PbI2 (s) + 2KNO3 (aq)

Read the question and identify the aim: finding the concentration of EXCESS LEAD IONS (Pb2+)

Now you ask yourself how can you get lead ions as a product of this reaction...

 

[Queenroot]

1. Write the equation as above
2. n = cV (find the moles)
3. Mystery step
4. Mystery step
5. Find new volume (0.023+0.077)
6. Use the C = n/V to find your lead concentration

 

[Sachin123]

Did you get 0.66M

 

[mohammy52]

Did you get 0.66M

Yep

 

[mohammy52]

Nice question

 

[siranchroller]

mohammy can u put up solution

 

[emilios]

Here's my working:

2KI + Pb(NO3)2 --> 2KNO3 + PBI2

n(KI) = Cv
= 1*0.023 mol
n(Pb(NO3)2) = 0.077 mol

Limiting reagant is the potassium iodide and from the formula, you need half as many mols of lead nitrate i.e. if we're using 0.023mols of KI, we're gonna need 0.0115 mols of lead.

Since 0.0115 mols of lead has been used, we're left with (0.077-0.0115) mols = 0.0655mols

Note that the volume that has been added is (0.023+0.077) L = 0.1L

So then C(lead ions) = n/v
= 0.0655/0.1
= 0.66 M