0,4 M NaOH + 0,2 M HCl

[Function]

Hi everyone

 

Rapid question:

 

Imagine a reaction between a 0,4 M NaOH-solution and a 0,2 M HCl-solution with equal volume. What are the concentrations then of Na(+), OH(-), Cl(-) and H(+) at equilibrium?

 

Tried to put up a table with initial quantity, difference and rest, but can't go further...

 

HCl + NaOH --> NaCl + H2O

0,2 0,4 0 0

0,2 0,2 0,2 0,2

0 0,2 0,2 0,2

 

Can anyone help me?

 

Thanks!

 

F.

 

EDIT:

 

I could sove it like this:

HCl --> H(+) + Cl(-)

0,2 0 0

0,2 0,2 0,2

0 0,2 0,2

 

NaOH --> Na(+) + OH(-)

0,4 0 0

0,4 0,4 0,4

0 0,4 0,4

 

0,2 moles of OH(-) could react with 0,2 moles of H(+) to form H2O.

0,2 moles of OH(-) left in a volume of 2 liters, so a concentration of 0,1 M OH(-).

0,4 moles of Na(+) left ... so 0,2 M Na(+)

0,2 moles of Cl(-) left ... so 0,1 M Cl(-).

This is also the correct answer.

 

Now, my problem is: I'd like to say that there's still OH(-) and H(+) in H2O, but why can't you .. you know? Bring them into account or however you may say it? So, I'd say you still have 0,4 moles of OH(-) and 0,2 moles of H(+)... Why not?

Edited June 27, 2014 by Function

[Iota]

Hi everyone

 

Rapid question:

 

Imagine a reaction between a 0,4 M NaOH-solution and a 0,2 M HCl-solution with equal volume. What are the concentrations then of Na(+), OH(-), Cl(-) and H(+) at equilibrium?

 

I'd assume that where you've said 'equilibrium' that you mean "reaction completion". If it were an equilibrium, you'd need a dissociation constant value in order to work out equilibrium concentrations.

Note that volumes are equal, therefore the molar ratio between NaOH and HCl is 2:1 respectively. Therefore you don't need to do anything to those values, you simply need a balanced equation, as you already have done.

 

 

0,2 moles of OH(-) could react with 0,2 moles of H(+) to form H2O.

0,2 moles of OH(-) left in a volume of 2 liters, so a concentration of 0,1 M OH(-). <--- where did you get 2 litres from?

0,4 moles of Na(+) left ... so 0,2 M Na(+)

0,2 moles of Cl(-) left ... so 0,1 M Cl(-).

This is also the correct answer.

 

There's 0.2M Cl- from the 0.2M NaCl produced.

 

There's 0.4M Na+ from the 0.2M NaOH not reacted, plus the 0.2M NaCl produced.

 

There's 0.2M OH- from the 0.2M NaOH not reacted.

 

Now, my problem is: I'd like to say that there's still OH(-) and H(+) in H2O, but why can't you .. you know? Bring them into account or however you may say it? So, I'd say you still have 0,4 moles of OH(-) and 0,2 moles of H(+)... Why not?

 

The reason you don't include water is because it does not behave ionically. That is, it does not dissociate into ions; it's covalent.

 

NaCl, HCl and NaOH are all ionic. HCl is a special case because it is actually covalently bonded, however it behaves ionically in solution.

[Enthalpy]

The reaction won't keep both H⁺ and OH^- because they react to form H₂O.. This is what happens in a neutralization - and nearly nothing else with strong acids and bases as in your example.

 

The product of H⁺ and OH^- concentrations in water is around 10^-14/L at room temperature (somewhat more because the neutralization heats the water) so both around 0,2M is impossible.