Some help with a couple chemistry problems?

[cougarblue84]

Hey everyone,

I'm probably missing some basic ideas, but I'm having a bit of trouble with the following problems. Could someone please help??

Q. At standard temperature and pressure, what volume of hydrogen will be produced when 6.5 g of zinc is dissolved in hydrochloric acid, as shown the reaction below?
Zn(s) + 2HCl(aq) ------> ZnCl2(s) + H2(g)

Q. What volume of 0.5 M NaOH is needed to neutralize 25 mL of 2 M H2SO4?

Q. CaF2 is added to a 0.1 M Ca(NO3)2 solution. At what concentration of F- will CaF2 begin to precipitate? (Ksp of CaF2 is 4 x 10-11)

Thanks for your help!!

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[drgreen]

1) @ STP P=1atm and RT=22.4
n=mass/M=6.5/65=.1 mol
use PV=nRT
(1)(V)=(.1)(22.4)
V=2.24L

2)N1V1=N2V2

N is normality which is (M)x(# of mol of useful base or acid equivalents)
N for NaOH=(.5)(1)=.5
N for H2SO4=(2)(2)=4

plug in...

(.5)(x)=(4)(25)
x=200 mL

3) Precipitate forms when a solution is saturated which occurs at Ksp

set up equations...

Ca(NO3)2 -> Ca2+ + 2NO3-
CaF2 -> Ca2+ + 2F-

let [Ca2+]=x therefore [2F-]=2x

given that our Ksp=4*10^-11
and Ksp=[Ca][F-]^2
4*10^-11=(x)(2x)^2

but [Ca] is not merely x. it also contains Ca ions from the dissociated .1M Ca(NO3)2 solution, looking at the equation, 1:1 molar ratio, we have .1M of Ca. So let's rewrite the formula.

4*10^-11=(x+.1)(2x)^2
look at the red "x", since our Ksp is so small, this "x" is neglegable compared to .1M, so we ignore it.

4*10^-11=(.1)(2x)^2
solving for x gives us 3.2*10^-6

since we know the concentration of F-, [F-]=2x

the concentration of F- that begins the precipitate is 6.2*10^-6

I hope this is all correct and is a help

 

[cougarblue84]

Thanks!

The first two were the correct answers. The third answer was 2E-5 in the answer key, but you had the right method. I think there was just a tiny error in the calculation at the end. Thanks so much for explaining those problems so well. I appreciate it.

 

[kkatranji]

For the third one, there was a long discussion about how to solve that same type of problem.

http://forums.studentdoctor.net/showthread.php?t=535897

harrygt had a really good explanation on how to approach them.

I think drgreen you made your mistake here.

given that our Ksp=4*10^-11
and Ksp=[Ca][F-]^2
4*10^-11=(x)(2x)^2

I believe the final equation should look like this...
4*10^-11=(.1)(x)^2
which equals 2E-5

I think the reasoning was because the F- is not coming from the CaF2 so you don't need to multiple the F- by 2 (since normally CaF2 would disocciate 2 moles of F) in the solubility equation.

hopefully harrgt can confirm if i'm right.

if I am I will be really happy/grateful.

 

[drgreen]

anytime