Question

Consider the following equilibrium system at 25°C and select the true statement below:

HA(aq) + H2O(l) ↔ H3O+(aq) + A−(aq)        Ka = 1.0 x 10−4 at 25°C

  

this is a product-favored system at 25°C

   

a solution of NaA(s) dissolved in pure water is predicted to have a pH < 7.00 at 25°C

  

(Kb of A−) = 1 / (Ka of HA)

  

A− is a stronger base than H2O

  

HA is a stronger acid than H3O+

Answer 1

What does an equilibrium constant tell

The values for equilibrium constants (K) a related to the concentration (amounts) of reactants and products in equilibrium mixtures. The "K" tells the equilibrium ratio of products to reactants.

The value for K is large when products dominate the mixture. This implies that products are 'favored' over reactants. When the value for K is small the reactants dominate the mixture and are 'favored' over products.

The expression for an equilibrium reaction is determined by the coefficients in the balanced equation.

When K is large i.e. K > 1 products are "favored" they dominate the mixture when K = 1 neither reactants nor products are favored. when K is small i.e. K < 1 reactants are "favored" they dominate the mixture

In the reaction that is in the question

HA(aq) + H₂O(l) ↔ H₃O⁺(aq) + A⁻(aq)        Ka = 1.0 x 10⁻⁴ at 25°C

We have K<1 so reactants are favoured so the statement "this is a product-favored system at 25°C" is wrong.

If you look at the reaction on the right side we have an acid being formed so as the reaction prodceeds to the right the pH of the mixture will become acidic that is pH<7.

So the statement "a solution of NaA(s) dissolved in pure water is predicted to have a pH < 7.00 at 25°C" is wrong because if you increase the concentration of A-(aq) the reaction will go from left to right and so H3O+ (aq) will be consumed so the pH will increase not decrease.

We are aware of the relation between Ka and Kb which is

by substitution and rearrangement, we obtain:

K_aK_b = K_w

Where the equation for the ionization of a weak base A¯ in solution is:

A¯ + H₂O HA + OH¯

and its K_b expression is:

	[HA] [OH¯]
Kb =	----------------
	[A¯]

So K_b = K_w/K_a

So (Kb of A−) = 1 / (Ka of HA) is wrong

If you look at the statement "A− is a stronger base than H₂O" This is true because if you look at the reaction on the right you have an acid and a base and on the left side you have a salt and water which are both neutral in terms of pH. So A^- has to be be a stronger base than water.

In keeping with this explaination the statement "HA is a stronger acid than H₃O⁺" is wrong because it is a salt and of neutral pH ans so it is a weaker in terms of acidity with respect to H₃O⁺(aq)

So the correct answer is The only statement true with respect to the equation is "A− is a stronger base than H2O"