Question

What concentration of ClO3– results when 637 mL of 0.583 M AgClO3 is mixed with 633 mL of 0.265 M Mn(ClO3)2?

Answer 1

Molarity = Moles/Litres

so,
0.637L of AgClO3
Dissociates into:
Ag(ClO3) -> Ag+ + ClO3-
Find moles of ClO3-


0.633mL of Mn(ClO3)2
Dissociates into
Mn(ClO3)2 -> Mn + 2(ClO3-)
Find moles of ClO3-


To find final concentration:
Volume of solution 1 + volume of solution 2 = New volume of solution

Moles of ClO3 in Solution 1 + Moles of ClO3 in solution 2 = Total moles of ClO3 in solution

Sub into,

Molarity = Moles/Liters

Solve for molarity and you have your answer

Answer 2

isn't there a concentration formal i could use?

Answer 3

Yes,

Molarity = moles/liters

Answer 4

you need to find the moles for both solutions individual and add their sum then you need to take the volumes of both solutions and add them together and then put them into the formula

Answer 5

To find moles you have volume of the original solution and concentration of the original solution sub into that formula and find moles. Then use the dissociation formula I provided to find moles of ClO3-

Answer 6

so for the first one,
633 mL = 0.633L
and Molarity = 0.583 M

thus,

0.583M = x/0.633L

solve for x to find moles of AgClo3 in 637 mL at a concentration of 0.583 M

Answer 7

i mean a formula like m1 * v1= m2*v2

Answer 8

n ohh okay

Answer 9

That is used for dilutions, this isn't necessarily a dilution, if anything we are equilibrating these two concentrations solutions (they will both be in between what they once were)

Answer 10

so the first x = 0.369

Answer 11

right

Answer 12

Now look at the formula for disassociation (this is an ionic compound),

Ag(ClO3) -> Ag+ + ClO3-

To go from moles of AgClO3 to ClO3-

you divide the number in front of the compound you are starting with, in this case it is 1
so,

0.369mol/1

Then you multiply that number by the number in front of ClO3-, in this case the number in front of ClO3- is 1, so,

(0.369mol/1)*1

I hope you understanding balancing of equations etc

Answer 13

okay and everything in the equation has a coefficient of 1 in front of the chemical formulas.

Answer 14

(0.369mol/1)*1 = moles of ClO3-
This is a case where for every 1 mole of AgClO3 for every mole of ClO3-

The next solution is not this case but just follow my steps and you will get the right answer

Answer 15

This is a case where for every 1 mole of AgClO3 there is 1 mole of ClO3- dissociated

Answer 16

oh okay n for the next part i got .168
so do add both variable now?

Answer 17

.369 +.167

Answer 18

Nope you made a mistake for calculating the moles in the second solution

Answer 19

you skipped a step

Answer 20

look at the dissociation of Mn(ClO3)2

Answer 21

it isn't 1 to 1

Answer 22

oh isnt
.265mole= x/.633?

Answer 23

that is correct but you just have moles of Mn(ClO3)2 at that point

Answer 24

you dont have moles of ClO3 in that particular solution

Answer 25

look at the dissociation formula for Mn(ClO3)2

Answer 26

what do you mean by dissociation formula sorry for the dumb question

Answer 27

Read through what I posted

Answer 28

It is very important you understand what I did for the first solution if you want to do well in chemistry

Answer 29

I posted the dissociation formula for it in my first post but you should understand how I got it

Answer 30

oh i see what your saying i thought ur talking bout a specific formula
so it dissociates into Mn^ +2 (ClO3-) right

but arent the coeffients still one?

Answer 31

then your compound would be
MnClO3+
since adding the charges,
2+ + -1 = +1

The actual compound is

Mn(ClO3)2

so,

Mn 2+ + ClO3- + ClO3-
so,
2+ + -1 + -1 = 0

Mn(ClO3)2
^
this means that there are two molecules of ClO3, hence the brackets with the subscript 2

Answer 32

so then it
.168/2 = .084 * 1= .084? right

Answer 33

okay thanks

Answer 34

Mn(ClO3)2 -> Mn 2+ + 2(ClO3-)

Answer 35

so there is a 1 in front of Mn(ClO3)2 and a 2 in front of ClO3-

Answer 36

divide by 1 and then multiply by 2

Answer 37

to get moles of ClO3-

Answer 38

make sure you understand this