[*GenChem*] 2 questions....!!Help!

[sminj85]

---

Members do not see this ad.

I got two simple questions. Can anyone help?
1. How many moles of AgIO3(Ksp=3.1*10^-8) will dissolve in one liter of a 10^-5 M solution of NaIO3?

2.A pot containing an unknow liquid is boiling on a stove. The temp. of the liquid is 102 Celcius, which of the floowing is not a possibility?
a. Liquid is solution of water and table salt.
b. The liquid is pure water and the stove is velow sea level.
c. The liquid is pure water and the stove is above sea level.
D The liquid is a solution of water and sugar.
E. The liquid is a mixture of water and antifreeze.

3. When do I use Normality? If the question asks about dilution, we do not care of N right? But I am confused when we have to use it !!


Thank you guys...🙂!
YOu all ahve a good night.

 

[Tejv24]

I'll give it a shot. Do you have the answer by any chance?

1) Answer: AgIO3 = 3.1 * 10^-3

** in my explanation i'll use 10 power as E power. In other words, 10^-5 can be read as E-5. (i know it's stupid, but wanted to be clear on that)

Aight, first thing first, eq.
Ksp = 3.1E-8 = [Ag3+] [IO3-]

Initially, before any AgIO3 dissolves the concentrations are
[Ag3+] = 0
[IO3] = E-5

@ Equilibrium: x M (AgIO3) dissolves, we get
[Ag3+] = x
[IO3] = x + E-5 {x from AgIO3 & E-5 from NaIO3}

Set up the eq.-
Ksp = 3.1E-8 = [Ag3+][IO3]
3.1E-8 = [x] [x+E-5] (in [IO3] x is assumed to be negligble sice Ksp is small)
x = 3.1E-3 mole/liter

so, there it is 0.0031 moles of AglO3 dissolves per liter of E-5 M NaIO3 solution.

Let me know if that helped alittle?

 

[Tejv24]

2) (C) Water boils at 100 C at sea level (apprx). @ higher altitudes (above sea level), lower pressure, thus water would boil at a lower temperature than 100 C.
A,D,E = all define colligative properties that results in boiling point elevation.

 

[sminj85]

here are the following choices.

a.sqrt 3.1x10^-8
b.(sqrt 3.1x10^-8)(E-5)
c.(sqrt 3.1x10^-8)/E-5
d.(E-5) -(sqrt 3.1x10^-8)
e. (sqrt 3.1x10^-8) - (E-5)

*sqrt = square root

I do not know where square roote came from.
I thought the answer should be just (3.1x10^-8)/(E-5) = 3.1 E-3..
Any ideas?

 

[Tejv24]

where you getting this question from?
I used common ion effect approach... ;-/ idk bro

 

[BenignDMD]

The answer is E.

AgIO3 dissociates into Ag+ and IO3- ions. The solubility of AgIO@ would then be:

ksp = [Ag+][IO3] = (X)(X) = X^2

3.1 x 10^-8 = X^2

sqrt 3.1 x 10^-8 = Solubility of AgIO3.

You already have a 10^-5 M NaIO3 so you could only add

sqrt (3.1 x 10^-8 M) - 10^-5 M

 

[STACM]

Can anyone explain how you go from:

3.1E-8 = [x] [x+10^-5]

to get:

sqrt (3.1 x 10^-8 M) - 10^-5 M


??? the math doesnt seem to stick

(BTW, this problem is from the Kaplan Online Test #59)

 

[151AND8TH]

Can anyone explain how you go from:

3.1E-8 = [x] [x+10^-5]

to get:

sqrt (3.1 x 10^-8 M) - 10^-5 M


??? the math doesnt seem to stick

(BTW, this problem is from the Kaplan Online Test #59)

This math doesnt make much sense to me neither..

for question 2.. I thought the answer would be B-- pure water below sea level

PV=nRT.. high pressure = high temperature

102 is above the normal bp of pure water at sea level -- hence high temperature

and high pressure is found below sea level.. is this correct>?

 

[klutzy1987]

Can anyone explain how you go from:

3.1E-8 = [x] [x+10^-5]

to get:

sqrt (3.1 x 10^-8 M) - 10^-5 M


??? the math doesnt seem to stick

(BTW, this problem is from the Kaplan Online Test #59)

We are dealing with Kaplan here. It is probably just another Kaplans mistake. The way TEJV24 outlined the problem is 100% correct and i would just assume that Kaplan is wrong once again. The answer should be 3.1*10^-3

 

[doc toothache]

1) Answer: AgIO3 = 3.1 * 10^-3

** in my explanation i'll use 10 power as E power. In other words, 10^-5 can be read as E-5. (i know it's stupid, but wanted to be clear on that)
Aight, first thing first, eq.
Ksp = 3.1E-8 = [Ag3+] [IO3-]
Initially, before any AgIO3 dissolves the concentrations are
[Ag3+] = 0
[IO3] = E-5
@ Equilibrium: x M (AgIO3) dissolves, we get
[Ag3+] = x
[IO3] = x + E-5 {x from AgIO3 & E-5 from NaIO3}
Set up the eq.-
Ksp = 3.1E-8 = [Ag3+][IO3]
3.1E-8 = [x] [x+E-5] (in [IO3] x is assumed to be negligble sice Ksp is small)
x = 3.1E-3 mole/liter
so, there it is 0.0031 moles of AglO3 dissolves per liter of E-5 M NaIO3 solution.

The contribution of IO3- at E-5 from NaIO3 is negligible leaving you with x. The expression then becomes x^2= 3.1 x 10^-8 and the [Ag+]=[IO3-]=[AgIO3]=sqrt 3.1 x 10^-8 (minus the 10^-5 IO3- which is already present in the solution) And as BenignDMD showed the answer is E.

We are dealing with Kaplan here. It is probably just another Kaplans mistake. The way TEJV24 outlined the problem is 100% correct and i would just assume that Kaplan is wrong once again. The answer should be 3.1*10^-3

Logically the answer at 3.1 x 10^-3 does not make sense. From the Ksp equation

[Ag+][IO3-]=3.1 x 10^-8, then
[Ag+]=[IO3]=[AgIO3]=(3.1 x 10^-8)^1/2 =1.76 x 10^-4, which is the max concentration of Ag+, IO3- or AgIO3 that will dissolve in water. If the answer is 3.1 x 10^-3 then what it means is that we have increased the solubility of AgIO3 in water by adding small amount of the common ion IO3-. By this account the solubility of a compound can be increased further by adding trace amount of IO3-, which does not happen. Most common ion effect problems deal with addition of an ion which is in greater concentration than the contribution from Ksp, making the latter (Ksp) contribution negligible. In this example it is just the opposite. The common ion (IO3-) is in a much lower concentration that the contribution from the Ksp, making it possible to ignore the original conc. of IO3-.