[Atome]

Hello everyone,

I arrived at the wrong answer for this following problem but can't see my mistake.

Could anyone guide me in the right direction?

Thank you.

---

26. Equal volumes of aqueuous solutions (each 1.0 M) of two substances, W and X, are mixed. The reaction below rapidly reaches equilibrium.

W_(aq) + X_(aq)  ::equil::Y _(aq) + Z_(aq)

At equiliribium, the concentration of Y is measured as 0.40 M. What is the value of K_c for the reaction above?

--
W_(aq)         +             X_(aq)                         Y _(aq)            +         Z_(aq)

1.0                           1.0                                    0                          0
-0.4                        -0.4                                   +0.4                     +0.4

= 0.6                       = 0.6                                = 0.4                    = 0.4

K_c = (0.4)²/(0.6)² = 0.44

However, the correct answer is (in white colour) 16.

[Borek]

Your initial concentrations are not 1. Reread the question.

[Atome]

Thanks for your response.

I see. The molarity of the solutions are each 1.0 M, but the volume of the solution is not 1.

Could you please explain how I would express the volume? Would it be 1.0 M(x L) so x mols?

I was hesitant to try this initially because then I would have two unknowns (x and K_c).

[leve]

Equal volumes of aqueuous solutions (each 1.0 M) of two substances, W and X, are mixed. The reaction below rapidly reaches equilibrium.

Imagine this, if you have x liters of 1.0 M A in a beaker and the same liters of 1.0 M B as A in another beaker and you add those two together, how does the volume change? How does the molarity of A and B change?

Remember that molarity is moles/L of solution

[Atome]

Thank you very much for your reply.

I answered the question, realising that the concentrations of both W and X become 0.5 M.