Chapter 14, Problem 129QRT

Interpretation Introduction

Interpretation:

The acids $HM,HQandHZ$ have to be arranged in order of decreasing strength.  The $K_a$ values of these acids have to be determined where possible.

Concept Introduction:

The conjugate base of a strong acid is weak.  The stronger base has the larger $pH.$

Answer to Problem 129QRT

The decreasing order of strength of acids $HM,HQandHZ$ is $HM>HQ>HZ.$  The $K_a$ values of these acids $HM,HQandHZ$ are $1\times 10^{-1},1\times 10^{-3}and1\times 10^{-5}$ respectively.

Explanation of Solution

Given that, the $pH$ values of sodium salts $NaM,NaQandNaZ$ are $7.0,8.0and9.0$ respectively.  The decreasing order of $pH$ values of these salts is given below.

  $NaM>NaQ>NaZ$

Thus, the decreasing order of the acidic strength of the conjugate acids $HM,HQandHZ$ is given below.

  $HM>HQ>HZ$

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

  $A^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons HA\left(aq\right)+O{H}^{-}\left(aq\right)$

Where, $A=M,QorZ$

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

  $K_b=\frac{\left[HA\right]\left[O{H}^{-}\right]}{\left[A^{-}\right]}$

A table can be set up as given below.

$\begin{array}{cccc}& A^{-}\left(aq\right)& HA\left(aq\right)& O{H}^{-}\left(aq\right)\\ Initialconc.\left(M\right)& 0.1& 0& 1.0\times {10}^{-7}\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 0.1-x& x& {10}^{-7}+x\end{array}$

The concentration of $O{H}^{-}$ comes from its autoionization that is $1.0\times {10}^{-7}.$

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

  $K_b=\frac{\left(x\right)\left({10}^{-7}+x\right)}{\left(0.1-x\right)}$

  $\begin{array}{c}1\times {10}^{-7}+x=\left[O{H}^{-}\right]={10}^{-pH}={10}^{-\left(14.00-pH\right)}={10}^{\left(pH-14.00\right)}\\ \\ x={10}^{\left(pH-14.00\right)}-1\times {10}^{-7}\end{array}$

For $NaZ$, $pH=9.0$

  $\begin{array}{l}x={10}^{\left(9.0-14.00\right)}-1\times {10}^{-7}=1\times {10}^{-5}-1\times {10}^{-7}=1\times {10}^{-5}\\ \\ K_b=\frac{\left(1\times {10}^{-5}\right)\left(1\times {10}^{-7}+1\times {10}^{-5}\right)}{\left(0.1-1\times {10}^{-5}\right)}=1\times {10}^{-9}\end{array}$

For $NaQ$, $pH=8.0$

  $\begin{array}{l}x={10}^{\left(8.0-14.00\right)}-1\times {10}^{-7}=1\times {10}^{-6}-1\times {10}^{-7}=1\times {10}^{-6}\\ \\ K_b=\frac{\left(1\times {10}^{-6}\right)\left(1\times {10}^{-7}+1\times {10}^{-6}\right)}{\left(0.1-1\times {10}^{-6}\right)}=1\times {10}^{-11}\end{array}$

For $NaM$, $pH=7.0$

  $x={10}^{\left(7.0-14.00\right)}-1\times {10}^{-7}=1\times {10}^{-7}-1\times {10}^{-7}=0$

This suggests that the reaction with water does not go toward products at all.  That means $K_{b}of{M}^{-}$ may be so small that the $pH$ of the solution is accounted for exclusively by the ionization of water.  If indeed the base does provide all the $O{H}^{-}$ ions to make the solution’s $pH$, then $\left[HM\right]=\left[O{H}^{-}\right]={10}^{-7.0}=1\times {10}^{-7}$:

  $K_b=\frac{\left(1\times {10}^{-7}\right)\left(1\times {10}^{-7}\right)}{\left(0.1\right)}=1\times {10}^{-13}$

For $HZ$, the value of $K_a$ can be calculated as shown below.

  $K_a=\frac{K_w}{K_b}=\frac{1\times {10}^{-14}}{1\times {10}^{-9}}=1\times {10}^{-5}$

Similarly, for $HQ$

  $K_a=\frac{K_w}{K_b}=\frac{1\times {10}^{-14}}{1\times {10}^{-11}}=1\times {10}^{-3}$

For $HM$

  $K_a=\frac{K_w}{K_b}=\frac{1\times {10}^{-14}}{1\times {10}^{-13}}=1\times {10}^{-1}$