Gen Chem questions

[dentaldoc1515]

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So I can't seem to figure out these two problems even though they seem really simple..if someone can please explain how to get the correct answer I'd really appreciate it 🙂

1. Concentrated sulfuric acid is 18M. What volume of water would be needed to dilute 25mL of concentrated H2SO4 solution to 3M?
a. 25mL
b. 50mL
c. 75mL
d. 125mL
e. 150mL

Answer: D


2. How many moles of AgIO3 (Ksp= 3.1 x 10^-8) will dissolve in one liter of a 10^-5M solution of NaIO3?
a. (√3.1 x 10^-8)
b. (√3.1 x 10^-8)(10^-5)
c.(√3.1 x 10^-8) / (10^-5)
d. (10^-5) - (√3.1 x 10^-8)
e. (√3.1 x 10^-8) - (10^-5)
**please note the square root is over the entire 3.1 x 10^-8**
Answer: E

 

[redchesus]

You start with: 18 M * 0.025 L = 0.45 mol H2SO4

0.45 mol / X L = 3 M

solve for X, X = 0.15 L = 150 mL

150 - 25 = 125 mL

I'll come back later for the second problem.

 

[UndergradGuy7]

#1 You could use M1 * V1 = M2* V2 and in this case you would get 150 mL, but you already have 25 mL so 150-25 = 125 mL.

For the second problem it is a solubility problem using the common ion effect. Normally you concentrations of AgIO3 would be Ksp = (Ag)(IO3). In this problem though, you have IO3 from NaIO3 so this effects how much of the AgIO3 will dissolve.

So, Ksp = (Ag)(IO3 from the AgIO3) = 3.1E-8 or (x)(x) = 3.1E-8 take the square root and you find out that x = 1.76E-4 M. 1.76E-4 is how much of each would dissolve if AgIO3 is the only thing present. But since we already have 10E-5 M of IO3 from NaIO3 the amount that will dissolve is 1.76E-4 - 10E-5.

 

[RSD2014]

So I can't seem to figure out these two problems even though they seem really simple..if someone can please explain how to get the correct answer I'd really appreciate it 🙂

1. Concentrated sulfuric acid is 18M. What volume of water would be needed to dilute 25mL of concentrated H2SO4 solution to 3M?
a. 25mL
b. 50mL
c. 75mL
d. 125mL
e. 150mL

Answer: D


2. How many moles of AgIO3 (Ksp= 3.1 x 10^-8) will dissolve in one liter of a 10^-5M solution of NaIO3?
a. (√3.1 x 10^-8)
b. (√3.1 x 10^-8)(10^-5)
c.(√3.1 x 10^-8) / (10^-5)
d. (10^-5) - (√3.1 x 10^-8)
e. (√3.1 x 10^-8) - (10^-5)
**please note the square root is over the entire 3.1 x 10^-8**
Answer: E

Problem#1: H2SO4 + H20 => diluted H2SO4
(25mL)(18M) + (X)(0) = (25mL+X)(3M)
(25mL)(18M) = (25mL+X)(3M)
solve for X.
X = 125mL

 

[dentaldoc1515]

I knew it was something small for number one that I was missing. Okay so for number two i don't get why its a minus and not a divided by. I thought that Ksp= (x)(x). so you're solving for one of the x's..so you take the square root of Ksp..then your left with (√3.1 x 10^-8)= (x)(10^-5)...don't you divide by 10^-5 to get x?

 

[UndergradGuy7]

I knew it was something small for number one that I was missing. Okay so for number two i don't get why its a minus and not a divided by. I thought that Ksp= (x)(x). so you're solving for one of the x's..so you take the square root of Ksp..then your left with (√3.1 x 10^-8)= (x)(10^-5)...don't you divide by 10^-5 to get x?

You do solve for one of the x's in the Ksp equation. Ksp = (x)(x). The x is the concentration of Ag from AgIO3. It is also the concentration of IO3 FROM the AgIO3. It tells you how much IO3 will dissolve. It does not take into account the IO3 from NaIO3.

Since NaIO3 dissolves and gives you IO3 (10E-5) then not as much can come from AgIO3. So you have to subtract.

Not really sure why you are setting this equal (√3.1 x 10^-8)= (x)(10^-5) ...

 

[dentaldoc1515]

OKay that makes sense.. I was misreading the problem. Thanks so much for the clarification.