Question

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equilibrium constant for the following reaction equal to? B(aq) + H+(aq) ⇌ HB+(aq)

Answer 1

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1:  $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2:  $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:  

$K_1=K_b$

$K_2=\frac{1}{K_w}$

Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

Hence, the value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Answer 2

Equilibrium constants are the depiction of the product and the reactant of the chemical reaction.  The value of the equilibrium constant for the reaction is $\rm \dfrac{K_{b}}{K_{w}}.$

What is equilibrium constant?

An equilibrium constant is the number of the amount of the product and the reactant involved in a chemical equilibrium reaction.

The reaction can be shown as:

$\rm K_{1}: \rm B + H_{2}O \leftrightharpoons HB^{+} + OH^{-}, \rm K_{b}$

$\rm K_{2}: H^{+} + OH^{-} \leftrightharpoons H_{2}O, \dfrac{1}{K_{w}}$

The net reaction is,

$\rm B + H^{+} \leftrightharpoons HB^{+}, K_{c}$

The equilibrium constant of the reaction will be given as the product of the first and second equilibrium constants.

The equilibrium constant for the net reaction:

$\rm K_{c} = K_{1} \times K_{2}$

Here, $\rm K_{1} = K_{b}$ and $\rm K_{2} = \dfrac{1}{K_{w}}$

Substituting values in the above equation:

$\begin{aligned}\rm K_{c} &= \rm K_{b}\times \dfrac{1}{K_{w}}\\\\&= \rm \dfrac{K_{b}}{K_{w}}\end{aligned}$

Therefore, the equilibrium constant for the reaction will be $\rm \dfrac{K_{b}}{K_{w}}.$

Learn more about the equilibrium constant here:

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