[greenmoon]

Hi,I'm new here.

I have problems finding the volumes of two solutions.

Two solutions react:
One is a solution of KOH molar concentration 0.35 M.
The other a solution of HNO3 molar concentration 0.35 M as well.
From the reaction of those two results a solution with the volume 2 L and pH=12.I need to find the initial volumes of the KOH and HNO3 solution.

[sjb]

Hi,I'm new here.

I have problems finding the volumes of two solutions.

Two solutions react:
One is a solution of KOH molar concentration 0.35 M.
The other a solution of HNO3 molar concentration 0.35 M as well.
From the reaction of those two results a solution with the volume 2 L and pH=12.I need to find the initial volumes of the KOH and HNO3 solution.

Can you write an equation for what happens when these solutions mix? What does the pH and volume tell you about the excess of one of the reagents?

[greenmoon]

Well the equation would be:
KOH + HNO3 -> KNO3 + H2O

Now if pH=12 that means the solution is a base.So pOH=2 therefore CM=10^-2 from which the number of moles is 2*10^-2=0.02.

To see which reagent is in excess I think I have to know the volumes of KOH and HNO3,calculate the number of moles and see which is bigger,thing which I don't know how to do.Is there another way to calculate the number of moles?

[fledarmus]

You reported your solutions as being 0.35 M. What does the M stand for? What is it's definition? That should help you get to the number of moles...

[greenmoon]

M stands for moles/liters.

So if it's 0.35 moles/1 liter then I'll get the number of moles by multiplying 0.35 by 2?So the number of moles would be 0.7 in each solution?Is it that simple? 

[fledarmus]

Not quite. Your final volume is 2 L, but you don't know yet how much of each solution your are using to get there. However, if you know what the concentration of OH^- is, you should be able to figure out where it came from, and how much of each solution you needed.

[greenmoon]

Since KOH -> K⁺+OH^-.That means in KOH there is one mole oh OH and 1 mole of K,so the concentration of
KOH is the same with the concentration of OH and is 0.02?

[Borek]

Since KOH -> K⁺+OH^-.That means in KOH there is one mole oh OH and 1 mole of K,so the concentration of KOH is the same with the concentration of OH and is 0.02?

Yes - just remember this is excess.

Assume you mixed V_KOH and V_HNO3.

Can you express the final volume (2 L) using these two unknowns?

Use the stoichiometry information to express the final concentration of excess KOH in terms of V_KOH and V_HNO3 and the final volume (2L).

This gives two equations in two unknowns.

[greenmoon]

Well since stoichiometrically 1 mole of KOH reacts with 1 mole of HNO3,if KOH=x then HNO3=x as well
I think since Volume=molar concentration*number of moles it would go like this:
0.35*number of moles_KOH+0.35*number of moles_HNO3=2
And the number of moles of KOH=x+0.02 while the number of moles of HNO3=x.
So I guess the final equation is:0.35*(x+0.02)+0.35x=2.
Did I get it right?

[Borek]

Volume=molar concentration*number of moles

No, it doesn't go this way. Please check.

[greenmoon]

Oops checked it's Volume=number of moles/molar concentration.

So then the equation is:
x+0.02/0.35+x/0.35=2 ?

[greenmoon]

Well for sure it's not done well because I've solved it and the volume of KOH would be 0.36 and of HNO3 0.34.And I have the answers and it's not that.Wonder what I didn't do well...

[fledarmus]

What are the units that you get? 0.36 and 0.34 what?

Just a tiny hint:

And the number of moles of KOH=x+0.02 while the number of moles of HNO3=x.

[greenmoon]

Well 0.36 and 0.34 moles.
Edited cause I meant moles.

[sjb]

Well 0.36 and 0.34 liters.

If that is the case, what is the total volume?