Author Topic: Chem help (Read 1075 times) Tweet Share

Chem · « **on:** January 09, 2013, 08:47:10 pm »

A solution is formed by mixing 50.0 mL of 1.0 potassium nitrate (KNO3) solution with 50.0 mL of 1.0 M calcium nitrate (Ca(NO3)2) solution. The concentration of nitrate ion (NO3-) in the resulting solution is?

REBORN · « **Reply #1 on:** January 09, 2013, 08:49:03 pm »

KNO3 -> K+ + NO3-

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

Can you go from here?

hint: find mol of nitrate ions (total), then find concentration.

b^3 · « **Reply #2 on:** January 09, 2013, 08:55:11 pm »

Very rusty on chem so we'll see how this goes.

Spoiler

The amount of $NO_{3}$ ions being provided by the potassium nitrate is
$n( NO_{3}^{-} )=c( KNO_{3}) \times v( KNO_{3} )$
$n( NO_{3}^{-} )=1.0 \times 0.050=0.050 \: \mathrm{mol}$
The amount of $NO_{3}$ ions being provided by the calcium nitrate is
$n( NO_{3}^{-} )=2c( Ca(NO_{3})_{2}) \times v(Ca(NO_{3})_{2})$
$n(NO_{3}^{-})=2\times1.0\times 0.050=0.100\: \mathrm{mol}$

So the total amount of $NO_{3}^{-}$ ions we have is 0.050+0.100=0.150 mol, So the concentration becomes
$c( NO_{3} ^{-} )= \frac{n( NO_{3} ^{-})}{v( NO_{3} ^{-})}=\frac{0.150}{0.050+0.050}=1.5\: \mathrm{M}$

EDIT: Ok LaTeX might be broken....

teletubbies_95 · « **Reply #3 on:** January 09, 2013, 09:22:26 pm »

hey thanks for the question!

n(KNO3) =n(KNO3)

therefore, 0.05mol

n(Ca(NO3)2)= 2x (No3)

= 0.1 mol

therefore = 0.05+0.1 / 0.05+0.05
=
1.5M

Hope that helped!

Lasercookie · « **Reply #4 on:** January 09, 2013, 09:32:59 pm »

Quote from: teletubbies_95 on January 09, 2013, 09:22:26 pm

therefore = 0.05+0.1 / 0.05+0.05
=
1/5M

did you make a small arithmetic or typing error here?

teletubbies_95 · « **Reply #5 on:** January 09, 2013, 11:14:49 pm »

sos typo ...better change that!

thanks!

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