Question

HBrO (aq) + H2O (l) ⇋ H3O+ (aq) + BrO- (aq)

Answer 1

Answer:

6.24 x 10-3 M

Explanation:

Hello,

In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:

$Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}$

Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change $x$, we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:

$5.2x10^{-5}=\frac{x*x}{0.749-x}$

Thus, we obtain a quadratic equation whose solution is:

$x_1=-0.00627M\\x_2=0.00624M$

Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.

Best regards.

Answer 2

Answer:

6.24×10¯³ M

Explanation:

Step 1:

The equation for the reaction is given below:

HBrO (aq) + H2O (l) ⇋ H3O+ (aq) + BrO- (aq)

Step 2:

The following data were obtained from the question:

Equilibrium constant, Kc = 5.2x10^-5

Concentration of HBrO, [HBrO] = 0.749 M

Concentration of BrO-, [BrO-] =?

Step 3:

Determination of the concentration of

BrO-.

We can obtain the concentration of BrO- as follow:

Initial concentration:

[HBrO] = 0.749 M

[H3O+] = 0

[BrO-] = 0

During reaction:

[HBrO] = - y

[H3O+] = +y

[BrO-] = +y

After reaction:

[HBrO] = 0.749 - y

[H3O+] = y

[BrO-] = y

Applying the equation constant equation:

Kc = [H3O+] [BrO-] / [HBrO]

5.2x10^-5 = y × y / 0.749

5.2x10^-5 = y² / 0.749

Cross multiply

y² = 5.2x10^-5 x 0.749

Take the square root of both side

y = √(5.2x10^-5 x 0.749)

y = 6.24×10¯³ M

[BrO-] = y = 6.24×10¯³ M

Therefore, the the concentration of BrO- at equilibrium is 6.24×10¯³ M