[getfirefox]

How do I work out the concentration of I^- ions in each of the 5 mixtures with differnt volumes in?

Thanks guys!

[sjb]

What is the total volume of your solutions? How much KI do you have initially?

[getfirefox]

What is the total volume of your solutions? How much KI do you have initially?

Its just a question in a book.

The total volume of mixture 1 would be 10cm³ wouldn't it (simply add across the table)?

Same for all of them.

[sjb]

At a simple level, yes, the volume is additive.

So how are the concentration and volume related, for a given amount of substance?

[getfirefox]

At a simple level, yes, the volume is additive.

So how are the concentration and volume related, for a given amount of substance?

I have no idea dude.

I'm given that the potassium iodide is at 1.00moldm³ and the volumes in the table?

How do I work out the concentration of the iodide ions?

[sjb]

OK, If you have 1.00 mol dm^-3, how many moles in 5.0 cm^-3?

[getfirefox]

OK, If you have 1.00 mol dm^-3, how many moles in 5.0 cm^-3?

5 moles?

[sjb]

No, not 5 mol. How many cm³ in 1 dm³ ?

edit typo

[getfirefox]

No, not 5 mol. How many cm³ in 1 dm³ ?

1000cm³

(edit, sorry, cut and paste)

[sjb]

OK, so if there are 1000 cm³ in 1 dm³, and you have 0.1 mol in 1 dm³, how many moles in 1000 cm³? In 5 cm³?

[getfirefox]

OK, so if there are 1000 cm³ in 1 dm³, and you have 0.1 mol in 1 dm³, how many moles in 1000 cm³? In 5 cm³?

So it would be:

1.00moldm^-3 / 1000

= 0.001 moles in 1000cm³

So in 5cm³, it would be

0.001mol x 5cm³

= 0.005moldm³ for Mixture 1 in the table.

Is that right (are the units right for the final answer too)?

[sjb]

The number is right for the initial solution (before dilution), but the units are not. mol dm^-3 x dm³ = ?