[iced]

Hey guys the question is

How many milligrams of MgI2  must be added to 240.0 mL of 8.38×10−2M KI   to produce a solution with [I-] = 0.1000 M?

The problem I am having is I do not know how to balance the equation. I keep thinking its MgI2 + KI -> Mg2+ + I- + K+. I am also unsure how to solve the problem. I know I can solve for the number of moles KI has but if I have a balanced equation and do the mole to mole ratio with [I-] I only can get Litres which wont be helpful.

Any help would be great. Thanks again.

[Borek]

There is no reaction so there is no reaction equation to balance. Just mass balance - calculate how much I^- you need, how much there is already present, how much must be added.

Edit: sorry, missed that crucial "no" in hurry.

[iced]

Sorry I am still a little confused. Can I use the 250mL to solve for the amount of moles there is of [I-]? Also would I not have to use the 8.38x10-2 M KI then, but then I could only get litres 

[Borek]

Can I use the 250mL to solve for the amount of moles there is of [I-]?

You have to! But first you have to decide whether it is 240 or 250.

Also would I not have to use the 8.38x10-2 M KI then, but then I could only get litres 

Now I am confused - do you know what M stands for? What is a concentration? Where did you get liters from?

[iced]

sorry I am just getting frustrated with this question. it is 240 mL. M= molarity= mol/L

that is why I thought you solve for mols of [I-] and then using mols you solve for volume of KI?

[Borek]

You solve for moles of I^-, thats correct.

No need to get frustrated, it is very simple.

You want to have 0.240×0.1000 moles of I^-, right?

At the moment you have 0.240×8.38×10⁻², right?

How much you have to add?

[iced]

I still not getting it. So I have 0.024 mols of [I-] and 0.0201 mols of KI. I am not sure what you maen by add

[Borek]

I still not getting it. So I have 0.024 mols of [I-] and 0.0201 mols of KI.

You are asked to prepare solution that WILL HAVE 0.0240 moles of I^-. You HAVE solution with 0.0201 moles of I^-. You know, like you need $240 to buy something, but you have only $201.

I am not sure what you maen by add

Have you read the question? It asks

How many milligrams of MgI2  must be added

[iced]

lol ok I was thinking that before but I thought it was too simple. so 0.024-0.0201 = 0.0039 mols


MgI2 Molar mass = 278.113 g/mol * 0.0039 mols?

[Borek]

Told you it was easy.

Just don't forget to convert to milligrams, and don't forget of significant digits.