Question

What is the concentration of nitrate ion in a 425 mL solution containing 32.0 g of Mg(NO3)2 (M= 148.3)?

Answer 1

The correct way to do these types of questions it to write and equation for the dissociation of the ionic compound and then find the moles of the ion.

Then use the formula for molarity: \(\sf Molarity=\dfrac{moles}{liters~of~solution}\)

Answer 2

How do I write an equation for the disassociation of the compound?

Answer 3

You have to do it based on the charges of the ions (nitrate ion is a polyatomic ion, one of several you should memorize).

\(\sf Mg(NO_3)_2\rightarrow Mg^{2+}+2~NO^-_3\)

Answer 4

So now I find the moles?

Answer 5

use the mass of the compound and the coefficients, just like you would if you had to find the theoretical yield of a product.

Answer 6

The mass of 2NO3 is 124.02

Answer 7

I'll simplify it for ya, find the moles of Mg(NO3)2, multiply them by two.

these are the moles of the nitrate ion, now divide this number by the volume of the solution (you have to convert it to liters first)

Answer 8

Okay so Mg(NO3)2 is 148.33 g/mol *2= 296.66
296.66/.425 = 698.02

Answer 9

Wait did I use the wrong amount? I should have used 32.0 g instead of 425 mL...right?

Answer 10

you have to first find the moles of Mg(NO3)2,

\(\sf moles=\dfrac{mass}{Molar~mass}\)

Answer 11

32g/ 148.33= .2157

Answer 12

now do the rest of what i said earlier.

...multiply by two. these are the moles of the nitrate ion, now divide this number by the volume of the solution (you have to convert it to liters first)

Answer 13

.2157*2= .4314
.4314/.425= 1.0151 M?

Answer 14

yup! thats it

Answer 15

Thank you for being patient!! I appreciate your help!

Answer 16

no problem! i'm glad you put in some effort, a lot of people just give up.