Dilution Question

[MDtoBe777]

---

Members do not see this ad.

Is there an easy way to solve these problems? I always get so confused.

Ex. Which of the following methods could be use to produce 20mL of 1.0 M solution of KI

1) mix 10mL each of 1M KNO3 and 1M AgI
2) Mix 20 ml of 1M KOH with I2
3) Mix 10 ml each of 2M HI and 2M KOH
4) Mix 10 ml each of 1M HI and 1M KOH

 

[FightingIrish04]

I think the answer is #4 - you just find the number of moles of KI in the final solution .2ml x 1.0 M = .02 moles KI . So 10 ml of 1M HI gives .01 moles of I and 10 ml of 1.0 M KOH gives .01 moles of K ( total of .02 moles ) in 20 ml of solution total = 1.0 M.

 

[bonez318ti]

mv=mv

you are looking to make a .02 L solution of 1 M KI..

so m*v = 1*.02 = .02

from the answers, you need to make an equivalent amount of KI.. but first you need the right solutes.. with KOH and HI, they both disasssociate in aqueous solution.. so you can get your K+ and I-
with choice D, the 1M solutions of KOH and HI will give you 1M of KI, but the combined volume is .02 L.. so you end up with same amount of KI as you are asked for.

 

[Tranceradu]

what's the reason that it can't be a)? is it because a precipitate will form (AgNo3)?

 

[has]

answer is (3).
20ml*1 = .02moles of KI which means .02 moles of K and .02 moles of I. .02 mole of K combines with .02 moles of I to form .02 moles of KI.
So, using 10 ml of 2M HI and 10 ml of 2M KNo3 would mean 10ml*2 = .02 moles each.

 

[shaq786]

darn...it is 3 lol

Can someone explain why the world of chemistry likes to to say .02 moles of I and .02 moles of K in the compound KI.

 

[bonez318ti]

answer is (3).
20ml*1 = .02moles of KI which means .02 moles of K and .02 moles of I. .02 mole of K combines with .02 moles of I to form .02 moles of KI.
So, using 10 ml of 2M HI and 10 ml of 2M KNo3 would mean 10ml*2 = .02 moles each.

10ml + 10ml = 20ml * 2M = .04 moles

D is correct (10ml + 10ml = 20ml * 1M = .02 moles)

 

[shaq786]

MDtoBE777 What is the answer????

 

[TiggidyTooth]

10ml + 10ml = 20ml * 2M = .04 moles

D is correct (10ml + 10ml = 20ml * 1M = .02 moles)

No, the answer must be C. You're forgetting that if you double the volume by adding 10ml of 1M solutes, you half the final concentration. D will give you a 0.5M solution and C gives you the desired 1M solution.

Tooth

 

[bonez318ti]

No, the answer must be C. You're forgetting that if you double the volume by adding 10ml of 1M solutes, you half the final concentration. D will give you a 0.5M solution and C gives you the desired 1M solution.

Tooth

Damn it, good call.

I stand corrected. C is correct.

 

[shaq786]

So now the question will all these practice exams help us cover all tricky ends when we take the real mcat? lol

 

[MDtoBe777]

The answer is 3.

 

[MDtoBe777]

answer is (3).
20ml*1 = .02moles of KI which means .02 moles of K and .02 moles of I. .02 mole of K combines with .02 moles of I to form .02 moles of KI.
So, using 10 ml of 2M HI and 10 ml of 2M KNo3 would mean 10ml*2 = .02 moles each.

Okay..so the first step is to realize that 20mL of a 1M solution of KI means 0.2 moles of K and 0.2 moles of I. Then look for the reagents that will provide you with those concentrations. Is this explanation right?

 

[Jonathan13180]

I agree, its (3), right? I think #4 would give you a .5M solution of each, and NOT 1M since you are doubling the volume added.

 

[eltrebor]

Yup the anser is (3). This is from practice test B of TPR's Practice test A-D, #9 of PS. Just to verify.