Chapter 18, Problem 73A

Interpretation Introduction

Interpretation:

The $k_b$ value of given base is to be calculated.

Concept introduction:

Base ionization constant tells about equilibrium that whether reactants are favored or products.

For a reaction,

$BOH\rightleftharpoons B^{+}+O{H}^{-}$

Base ionization constant is given by:

$\begin{array}{l}{k}_b=\frac{\left[B^{+}\right]\times \left[O{H}^{-}\right]}{\left[BOH\right]}\\ where,\\ \left[B^{+}\right]isconcentrationof{B}^{+}\\ \left[O{H}^{-}\right]isconcentrationofO{H}^{-}\\ \left[BOH\right]isconcentrationofBOH\end{array}$

Answer to Problem 73A

$k_b=7.2\times {10}^{-7}$

Explanation of Solution

Ionization equation is given by:

$Za{H}_2+H_{2}O\rightleftharpoons O{H}^{-}+Za{H}_3{}^{+}$

Given data:

$Atequilibrium,\left[Za{H}_2\right]=0.0997M$

$\left[O{H}^{-}\right]=2.68\times {10}^{-4}mol/L$

Now $\left[Za{H}_3{}^{+}\right]=\left[O{H}^{-}\right]$

Therefore,

$\begin{array}{c}{k}_b=\frac{2.68\times {10}^{-4}\times 2.68\times {10}^{-4}}{0.0997}\\ =7.2\times {10}^{-7}\end{array}$

Conclusion

The value of $k_b$ of base is $k_b=7.2\times {10}^{-7}$