Lets say you pipet into a 250 mL volumetric flask the following:

 

10 ml of 3e-4 M bromscresol green solution

25 mL of 1.6 M acetic acid (HC2H3O2)

10 mL of .200 M KC1 Solution

 

and diluted to the 250 mL mark

 

What is the molarity of acetic acid in this solution?

 

Do you calculate the number of moles in the 25 mL solution since you know the molarity, then when u get the moles you find molarity over 250 mL?

 

2nd problem.. Lets say you have a solution with the following:

 

10 mL of 3e-4 Bromscresol green solution

10 mL of .160 M sodium acetate solution

and diluted to 250 mL mark

 

What is the mles acetate present?

 

Do you calculate the # of moles over just 10 mL?

 

Does the # of moles stay constant even though you add more volume? I think i heard that somewhere but not sure

it`s all percentages, 1 mole in 100ml of soln, will still contain the one mole, but if you take 1ml from it. then will have only 1/100`th of a mole in that sample.

and so your 25 ml in 250, we may say 10 in 100 (or 1 tenth dillution), and you can work from there, the bromocresol green you can discount as part of the reaction as that`s only the indicator and should already come pre-buffered.

the molarity of the solution is equivalent to moles/liters.

 

.04 Moles / total volume in liters = .04/.250 L = .16 M

 

When determining molarity for this problem, entire volume must be used. Our volume is 250 mL, convert that to Liters and receive .250 L. Yes, u take the # of moles of the initial acetic acid solution.

 

Part 2, the number of moles always stays the same; it is not affected by changes in volume. The number of moles is .0016 before and after.