Question

beaker A contains 0.4M NaOH and beaker B contains 0.2M HCl both have equal volumes. Some one mixes the contents, what are the concentrations of the compounds in the solution at equilibrium?

Answer 1

@Callisto

Answer 2

@experimentX

Answer 3

can you help?

Answer 4

this is a neutralisation reaction
HCl+NaOH >> H2O+NaCl
H+ + OH- >> H2O

let the volume of each be 1 L
(since the concentration is calculated in moles per litre)

n(HCl) before reaction: 0.2M n(NaOH) before: 0.4M

from the neutralisation equation
we know that 1mole of HCl reacts with 1 mole of NaOH
so after the reaction:
n(HCl)= none
n(H+) = none (all reacted)
n(Cl-) = 0.2mol/2L(total volume of solution)
= 0.1 M (Cl- are spectator ions so the Cl- ions are not consume in the reaction)
n(NaOH)= 0.2 mol/ 2L
= 0.1 M ( the other 0.2 mole is neutralise with the H+ from HCl)
n(Na+) = 0.4 mol/ 2L
= 0.2 M (Na+ ions are also spectator ions, so the n of moles remain unchanged)
n(OH-) = n(NaOH)
= 0.1M