At 25°C, the equilibrium constant Kc for the reaction 2A(aq) ↔ B(aq) + C(aq) is 65. If 2.50 mol of A is added to enough water to prepare 1.00 L of solution, what will the equilibrium concentration of A be?

0.146 M

Explanation:

Equation for the reaction :

2A(aq) ↔ B(aq) + C(aq)

= 65

Molar concentration of A =

= 2.5 M

                       2A(aq)     ↔        B(aq)     +   C(aq)

Initial              2.50                      0                0

Change          - 2x                      + x             + x

Equilibrium   2.50 - 2x               +x               +x

8.062(2.5 - 2x) = x

20.155 - 16.124x = x

20.155 = 16.124x+x

20.155 = 17.124x

x =

x = 1.177

[A] = 2.5 - 2x

= 2.5 - 2(1.177)

= 0.146 M

Therefore, the equilibrium concentration of A = 0.146 M

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.