Chapter 14, Problem 121QRT

Interpretation Introduction

Interpretation:

The $pHof0.00500MN{H}_4\left[B{\left(OH\right)}_4\right]$ has to be calculated (a) by using the simplifying assumption regarding x, (b) by solving a quadratic equation and (c) by using the method of successive approximations.

Answer to Problem 121QRT

The $pHof0.00500MN{H}_4\left[B{\left(OH\right)}_4\right]$ (a) by using the simplifying assumption regarding x is $10.46$, (b) by solving a quadratic equation is $10.45$ and (c) by using the method of successive approximations is $10.45.$

Explanation of Solution

The solution contains $N{H}_4{}^{+}$ and ${\left[B{\left(OH\right)}_4\right]}^{-}$ ions.  $K_a$ for $N{H}_4{}^{+}$ is $5.6\times {10}^{-10}$ and $K_b$ for ${\left[B{\left(OH\right)}_4\right]}^{-}$ is $1.7\times {10}^{-5}.$  As $K_b>K_a$, the solution is basic.

The equation for the equilibrium can be written as given below.

  $B{\left(OH\right)}_4{}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons B{\left(OH\right)}_3{H}_{2}O\left(aq\right)+O{H}^{-}\left(aq\right)$

The basic ionization constant for the above equation can be written as given below.

  $K_b=\frac{\left[B{\left(OH\right)}_3\left(H_{2}O\right)\right]\left[O{H}^{-}\right]}{\left[B{\left(OH\right)}_3{}^{-}\right]}$

A table can be set up as given below.

$\begin{array}{cccc}& B{\left(OH\right)}_4{}^{-}\left(aq\right)& B{\left(OH\right)}_3{H}_{2}O\left(aq\right)& O{H}^{-}\left(aq\right)\\ Initialconc.\left(M\right)& 0.00500& 0& 1.0\times {10}^{-7}\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 0.00500-x& x& x\end{array}$

The concentration of $O{H}^{-}$ comes from its auto ionization that is $1.0\times {10}^{-7}.$  It can be assumed to be small compared to the change.

By substituting all the values in the above equation, the value of $x$ can be calculated.

  $K_b=\frac{\left(x\right)\left(x\right)}{\left(0.00500-x\right)}=1.7\times {10}^{-5}$

(a) Assuming x is very small, it can be written as $0.00500-x\cong \mathrm{0.00500.}$

Then,

  $\begin{array}{l}{x}^2=\left(1.7\times {10}^{-5}\right)\left(0.00500\right)\\ \\ x=2.9\times {10}^{-4}M=\left[O{H}^{-}\right]\end{array}$

The $pH$ of the solution can be calculated as follows,

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(2.9\times {10}^{-4}\right)=\mathrm{3.54.}\\ \\ pH=14.00-pOH=14.00-3.54=\mathrm{10.46.}\end{array}$

Therefore, the $pHof0.00500MN{H}_4\left[B{\left(OH\right)}_4\right]$ is $10.46.$

(b) The above equation can be rearranged in the form of a quadratic equation as shown below.

  $\begin{array}{c}{x}^2=\left(1.7\times {10}^{-5}\right)\left(0.00500-x\right)\\ \\ x^2+\left(1.7\times {10}^{-5}\right)x-8.5\times {10}^{-8}=0\\ \\ x=2.8\times {10}^{-4}=\left[O{H}^{-}\right]\end{array}$

The $pH$ of the solution can be calculated as follows,

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(2.8\times {10}^{-4}\right)=\mathrm{3.55.}\\ \\ pH=14.00-pOH=14.00-3.55=\mathrm{10.45.}\end{array}$

Therefore, the $pHof0.00500MN{H}_4\left[B{\left(OH\right)}_4\right]$ is $10.45.$

(c) using the method of successive approximations, it can be written as given below.

  $0.00500-x=0.00500-2.9\times {10}^{-4}M=0.00471M.$

Now,

  $\begin{array}{l}{x}^2=\left(1.7\times {10}^{-5}\right)\left(0.00471\right)\\ \\ x=2.8\times {10}^{-4}M=\left[O{H}^{-}\right]\end{array}$

The $pH$ of the solution can be calculated as follows,

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(2.8\times {10}^{-4}\right)=\mathrm{3.55.}\\ \\ pH=14.00-pOH=14.00-3.55=\mathrm{10.45.}\end{array}$

Therefore, the $pHof0.00500MN{H}_4\left[B{\left(OH\right)}_4\right]$ is $10.45.$