Function Posted June 27, 2014 Share Posted June 27, 2014 (edited) 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 Link to comment Share on other sites More sharing options...
Iota Posted June 28, 2014 Share Posted June 28, 2014 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. Link to comment Share on other sites More sharing options...
Enthalpy Posted June 28, 2014 Share Posted June 28, 2014 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. Link to comment Share on other sites More sharing options...
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