Chapter 17, Problem 91AP

A solution is made by mixing exactly 500 mL of 0.167 M NaOH with exactly 500 mL 0.100 M HCOOH. Calculate the equilibrium concentrations of ${\text{H}}_{\text{3}}{\text{O}}^{\text{-}}{\text{, HCOOH, HCOO}}^{\text{-}}{\text{, OH}}^{\text{-}}{\text{, and Na}}^{\text{+}}\text{.}$

Interpretation Introduction

Interpretation:

Theequilibrium concentrations of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$, $\text{HCOOH}$, ${\text{HCOO}}^{-}$, ${\text{OH}}^{\text{-}},$ and ${\text{Na}}^{\text{+}}$ in a solution made by mixing given volume of $0.167\text{M NaOH}$ with given volume of $0.100\text{M HCOOH,}$ are to be calculated.

Concept introduction:

The $\text{pH}$ measures the concentration of hydronium ions in a solution. The solution with ahigh concentration of hydronium hasa low $\text{pH}$ value and the solution with a low concentration of hydronium ions hasa high $\text{pH}$ value.

If an anion reacts with water, it is called anionic hydrolysis. If a cation reacts with water, it is called cationic hydrolysis

Answer to Problem 91AP

Solution: The equilibrium concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\text{ is 3}\times {\text{10}}^{-3}\text{M}$, the equilibrium concentration of $\left[\text{HCOOH}\right]\text{ is }8.8\times {10}^{-11}\text{ M}$, the equilibrium concentration of $\left[{\text{HCOO}}^{-}\right]\text{ is }0.0500\text{ M}$, the equilibrium concentration of $\left[{\text{OH}}^{-}\right]\text{ is }0.0335\text{M}$, and the equilibrium concentration $\left[{\text{Na}}^{+}\right]\text{ is }0.0835\text{M}$.

Explanation of Solution

Given information: The concentration of $\text{NaOH}$ is $0.167\text{M }$.

The volume of $\text{NaOH}$ is $500\text{mL}$ or $0.500\text{L}$.

The concentration of $\text{HCOOH}$ is $0.100\text{M }$.

The volume of $\text{HCOOH}$ is $500\text{mL}$ or $0.500\text{L}$.

Moles of $\text{NaOH}$ ions is calculated using expression as follows:

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of sol}\text{.}$

Substitute $0.167\text{M }$ for theconcentration of $\text{NaOH}$ and $0.500\text{L}$ for the volume in the above expression as follows:

$\begin{array}{l}\text{Moles of NaOH}=0.167M\times 0.500L\\ \text{ =}\frac{0.167\text{mol}}{1\text{L sol}}\times 0.500\text{L}\\ =0.0835\text{mole}\text{.}\end{array}$

Moles of $\text{HCOOH}$ ions is calculated using the expression as follows:

$\text{Moles of HCOOH}=\text{Concentration of HCOOH}\times \text{Volume of sol}$

Substitute $0.100\text{M }$ for the concentration of $\text{HCOOH}$ and $0.500\text{L}$ for the volume in the above expression as follows:

$\begin{array}{l}\text{Moles of HCOOH}=0.100M\times 0.500L\\ \text{ =}\frac{0.100\text{mol}}{1\text{L sol}}\times 0.500\text{L}\\ =0.0500\text{mole}\text{.}\end{array}$

Summarize the concentration at equilibrium as follows:

$\begin{array}{l}\text{HCOOH}\left(aq\right)+\text{NaOH}\left(aq\right)\to \text{HCOONa}\left(aq\right).\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& 0.0500& 0.0835& 0\\ \text{Change}\left(\text{M}\right)& -0.0500& -0.0500& +0.0500\\ \text{Equilibrium}\left(\text{M}\right)& 0& 0.0355& 0.0500\end{array}\end{array}$

The resulting solution is not a buffer solution.

The volume of the resulting solutionis as follows:

$\begin{array}{c}\left(\text{500mL+500mL}\right)=\text{1000 mL}\\ =\text{1L}\text{.}\end{array}$

Concentration of $\left[{\text{Na}}^{+}\right]$ is calculated using the expression as follows:

$\text{Concentration}\text{of}\left[{\text{Na}}^{+}\right]=\frac{\text{Moles}\text{of}{\text{Na}}^{+}}{\text{Volume}\text{of}\text{sol}}$.

Substitute $\left(0.0335+0.0500\right)\text{mol}$ for the moles of ${\text{Na}}^{+}$ and $\text{1L}$ for the volume in the above expression as follows:

$\begin{array}{l}\text{Concentration}\text{of}\left[{\text{Na}}^{+}\right]=\frac{\left(0.0335+0.0500\right)\text{mol}}{1\text{L}}\\ =0.0835\text{M}\text{.}\end{array}$

Concentration of $\left[{\text{HCOO}}^{-}\right]$ is calculated using the expression as follows:

$\text{Concentration}\text{of}\left[{\text{HCOO}}^{-}\right]=\frac{\text{Moles}\text{of}{\text{HCOO}}^{-}}{\text{Volume}\text{of}\text{sol}}$.

Substitute, $0.0500\text{mol}$ for the moles of ${\text{HCOO}}^{-}$ and $\text{1L}$ for the volume in the above expression as follows:

$\begin{array}{c}\text{Concentration}\text{of}\left[{\text{HCOO}}^{-}\right]=\left(\frac{0.0500\text{mol}}{1\text{L}}\right)\\ =0.0500\text{M}\text{.}\end{array}$

Concentration of $\left[{\text{OH}}^{-}\right]$ is calculated using the expression as follows:

$\text{Concentration}\text{of}\left[{\text{OH}}^{-}\right]=\frac{\text{Moles}\text{of}{\text{OH}}^{-}}{\text{Volume}\text{of}\text{sol}}$.

Substitute $0.0335\text{mol}$ for the moles of ${\text{OH}}^{-}$ and $\text{1L}$ for the volume in the above expression as follows:

$\begin{array}{c}\text{Concentration}\text{of}\left[{\text{OH}}^{-}\right]=\left(\frac{0.0335\text{mol}}{1\text{L}}\right)\\ =0.0335\text{M}\end{array}$.

Concentration of $\left[{\text{H}}^{+}\right]$ is calculated using the expression as follows:

$\text{Concentration}\text{of}\left[{\text{H}}^{+}\right]\times \text{Concentration}\text{of}\left[{\text{OH}}^{-}\right]={10}^{-14}$.

Substitute $0.0335\text{M}$ for the concentration of ${\text{OH}}^{-}$ in the above expression as follows:

$\begin{array}{l}\text{Concentration}\text{of}\left[{\text{H}}^{+}\right]=\frac{1\times {10}^{-14}}{0.0335}\\ =3\times {10}^{-13}\text{M}\text{.}\end{array}$

The concentration of $\left[{\text{HCOO}}^{-}\right]$ is calculatedas follows:

${\text{HCOO}}^{-}$ is a weak base. The conjugate base undergoes anionic hydrolysis

Summarize the concentration at equilibrium as follows.

Consider $x$ to be a degree of dissociation.

$\begin{array}{l}{\text{HCOO}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons \text{HCOONa}\left(aq\right)+{\text{OH}}^{-}\left(aq\right).\\ \begin{array}{ccccc}\text{Initial}\left(\text{M}\right)& 0.0500& -& 0& 0.0335\\ \text{Change}\left(\text{M}\right)& -x& -& +x& +x\\ \text{Equilibrium}\left(\text{M}\right)& 0.0500-x& -& x& 0.0335+x\end{array}\end{array}$

The equilibrium expression for a reaction is written as follows:

${\text{K}}_{\text{b}}=\frac{\left[\text{HCOONa}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{HCOO}}^{-}\right]}.$

Here, ${\text{K}}_{\text{b}}$ is the base dissociation constant, $\left[{\text{HCOO}}^{-}\right]$ is concentration of methanoate ions, $\left[{\text{OH}}^{-}\right]$ is the concentration of hydroxide ions, and $\left[\text{HCOONa}\right]$ is the concentration of sodium formate.

Substitute $\left(0.0500-x\right)$ for the concentration of $\left[{\text{HCOO}}^{-}\right]$, $\left(0.0335+x\right)$ for the concentration of $\left[{\text{OH}}^{-}\right]$, $\left(x\right)$ for the concentration of $\left[\text{HCOONa}\right]$, and $5.9\times {10}^{-11}$ for base dissociation constant ${\text{K}}_{\text{b}}$ in the above expression as follows:

$5.9\times {10}^{-11}=\frac{\left(x\right)\left(0.0335+x\right)}{\left(0.0500-x\right)}$.

The value of x is very small as compared to $\text{0}\text{.0335 and 0}\text{.0500}$. It can be neglected.

$\begin{array}{l}5.9\times {10}^{-11}=\frac{\left(x\right)\left(0.0335\right)}{\left(0.0500\right)}\\ x=8.8\times {10}^{-11}\text{M}\text{.}\end{array}$

The concentration of $\left[\text{HCOOH}\right]=x=8.8\times {10}^{-11}\text{M}$.

Conclusion

The equilibrium concentrations are as follows:

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]=\text{ 3}\times {\text{10}}^{-3}\text{M}\\ \left[\text{HCOOH}\right]=8.8\times {10}^{-11}\text{M}\\ \left[{\text{HCOO}}^{-}\right]=0.0500\text{M}\\ \left[{\text{OH}}^{-}\right]=0.0335\text{M}\\ \left[{\text{Na}}^{+}\right]=0.0835\text{M}\text{.}\end{array}$