Chapter 5, Problem 5.37P

Interpretation Introduction

Interpretation:

The $pH$ curve for the titration of $25.0c{m}^3$ of $0.15MBa{\left(OH\right)}_{2}with0.22MHCl$ has to be drawn. 

Concept Introduction:

For the titration of a weak base with a strong acid, the $pH$ before the equivalence point depends on the excess concentration of base and the $pH$ after the equivalence point depends on the excess concentration of acid.  At the equivalence point, there is not an excess of either acid or base so $\text{pH}$ is $\text{7}\text{.0}$.  The equivalence point occurs when total volume of base has been added. 

$pH=-\log \left[H_3{O}^{+}\right]$

$pH=-\log \left[O{H}^{-}\right]$

$pH+pOH=14$

Calculation of Moles:

1. $No.ofmoles\left(n\right)=\frac{W}{M.W}$

Where,

$\begin{array}{l}W =Weightofagivencompound\\ \\ M.W=Molecularweightofthesamecompound\end{array}$

2. $No.ofmoles\left(n\right)=\left(Molarity\right)\left( Volume\right)$

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

$Molarity=\frac{\text{moles of solute}}{\text{volume of solution in L}}$

Explanation of Solution

The reaction occurring in the titration is the neutralization of $O{H}^{-}$ (from $Ba{\left(OH\right)}_2$) by $H_3{O}^{+}$ (from $HCl$):

$\begin{array}{l}Ba{\left(OH\right)}_2\left(aq\right)+2HCl\left(aq\right)\to 2H_{2}O\left(l\right)+BaC{l}_2\left(aq\right)\\ \\ H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)\end{array}$

Which shows that $2$ equivalent of acid react with $1$ equivalent of Barium oxide. For achieving equivalence point, we need the equivalent quantities of acid and base.  So

$M_1{V}_1=M_2{V}_2$

Where,

$M_1=$ Concentration of base

$M_2=$ Concentration of acid

$V_1=$ Volume of base

$V_2=$ Volume of acid

Substituting the given values in the above equation and solving for $V_2$,

$\begin{array}{l}{M}_1{V}_1=M_2{V}_2\\ \\ \left(0.15M\right)\left(25mL\right)=\left(2\times 0.22\right)\left(V_2\right)\\ \\ V_2=\frac{\left(0.15M\right)\left(25mL\right)}{2\times 0.22}=8.522mL.\end{array}$

 For neutralization we need $8.522mL$ of acid. 

For plotting the graph we can start the titration by taking $0.00mL$ of acid and gradually add acid in dropwise manner till $8.5mL$. 

Case – 1:

Given data:

$\begin{array}{l}Volumeof Ba{\left(OH\right)}_2 =\text{ 25}.00ml=25.00\times {10}^{-3}L\\ \\ Concentrationof Ba{\left(OH\right)}_2 =0.15M\\ \\ ConcentrationofHCl=0.22M\\ \\ VolumeofHCladded=0.00ml=0.00\times {10}^{-3}L\\ \\ Thetotalvolumeofthesolutionatthispo\mathrm{int}=25.00\times {10}^{-3}L+0.00\times {10}^{-3}L\\ \\ =25.00\times {10}^{-3}L\end{array}$

$\begin{array}{l}TheinitialnumberofmolesofBa{\left(OH\right)}_2=Molarity\times Volume\\ \\ =\left(0.15M\right)\left(25.00\times {10}^{-3}L\right)\\ \\ =3.75\times {10}^{-3}.\end{array}$

$\begin{array}{l}MolesofaddedHCl=Molarity\times Volume\\ \\ =\left(0.22M\right)\left(0.00\times {10}^{-3}L\right)\\ \\ =0.\end{array}$

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{No.ofmolesBa{\left(OH\right)}_2\left(aq\right)+2HCl\left(aq\right)\to 2H_{2}O\left(l\right)+BaC{l}_2\left(aq\right)}\\ Initial3.75\times {10}^{-3}0-0\\ Change00-0\end{array}}\\ Final3.75\times {10}^{-3}0-0\end{array}$

$\begin{array}{l}Excess\left[O{H}^{-}\right]=\frac{3.75\times {10}^{-3}moles}{25.00\times {10}^{-3}L}\\ \\ =0.15M.\end{array}$

$\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]\\ \\ =-\log \left(0.15\right)\\ \\ =\mathrm{0.8239.}\end{array}$

$\begin{array}{l}pH=14-pOH\\ \\ =14-0.8239=\mathrm{13.1761.}\end{array}$

Therefore, the initial $pH$ of the solution has been calculated to be $13.1761$. 

Case – 2:

$\begin{array}{l}VolumeofHCladded=5.00ml=5.00\times {10}^{-3}L\\ \\ Thetotalvolumeofthesolutionatthispo\mathrm{int}=25.00\times {10}^{-3}L+5.00\times {10}^{-3}L\\ \\ =30.00\times {10}^{-3}L\end{array}$

$\begin{array}{l}MolesofaddedHCl=Molarity\times Volume\\ \\ =\left(0.22M\right)\left(5.00\times {10}^{-3}L\right)\\ \\ =1.1\times {10}^{-3}.\end{array}$

As there is two molecules of $HCl$, moles of added $HCl$ will be $2.2\times {10}^{-3}$. 

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{No.ofmolesBa{\left(OH\right)}_2\left(aq\right)+2HCl\left(aq\right)\to 2H_{2}O\left(l\right)+BaC{l}_2\left(aq\right)}\\ Initial3.75\times {10}^{-3}2.2\times {10}^{-3}-0\\ Change-2.2\times {10}^{-3}-2.2\times {10}^{-3}-+2.2\times {10}^{-3}\end{array}}\\ Final1.55\times {10}^{-3}0-2.2\times {10}^{-3}\end{array}$

$\begin{array}{l}Excess\left[O{H}^{-}\right]=\frac{1.55\times {10}^{-3}moles}{30.00\times {10}^{-3}L}\\ \\ =0.0516M.\end{array}$

$\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]\\ \\ =-\log \left(0.0561\right)\\ \\ =\mathrm{1.2873.}\end{array}$

$\begin{array}{l}pH=14-pOH\\ \\ =14-1.2873=\mathrm{12.7127.}\end{array}$

Therefore, the $pH$ of the solution has been calculated to be $\text{12}\text{.7127}$. 

Case – 3:

$\begin{array}{l}VolumeofHCladded=8.522ml=8.522\times {10}^{-3}L\\ \\ Thetotalvolumeofthesolutionatthispo\mathrm{int}=25.00\times {10}^{-3}L+8.522\times {10}^{-3}L\\ \\ =33.522\times {10}^{-3}L\end{array}$

$\begin{array}{l}MolesofaddedHCl=Molarity\times Volume\\ \\ =\left(0.22M\right)\left(8.522\times {10}^{-3}L\right)\\ \\ =1.8748\times {10}^{-3}.\end{array}$

As there is two molecules of $HCl$, moles of added $HCl$ will be $3.75\times {10}^{-3}$. 

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{No.ofmolesBa{\left(OH\right)}_2\left(aq\right)+2HCl\left(aq\right)\to 2H_{2}O\left(l\right)+BaC{l}_2\left(aq\right)}\\ Initial3.75\times {10}^{-3}3.75\times {10}^{-3}-0\\ Change-3.75\times {10}^{-3}-3.75\times {10}^{-3}-+3.75\times {10}^{-3}\end{array}}\\ Final00-3.75\times {10}^{-3}\end{array}$

The $HCl$ will react with an equal amount of the $Ba{\left(OH\right)}_2$ and $\text{0}\text{.0 mol}$ $Ba{\left(OH\right)}_2$ will remain.  This is the equivalence point of a weak base –strong acid titration, thus, the $pH$ is $7.0$.  Only the neutral salt $BaC{l}_2$ is in solution at the equivalence point. 

Case – 4:

$\begin{array}{l}VolumeofHCladded=10.00ml=10.00\times {10}^{-3}L\\ \\ Thetotalvolumeofthesolutionatthispo\mathrm{int}=25.00\times {10}^{-3}L+10.00\times {10}^{-3}L\\ \\ =35.00\times {10}^{-3}L\end{array}$

$\begin{array}{l}MolesofaddedHCl=Molarity\times Volume\\ \\ =\left(0.22M\right)\left(10.00\times {10}^{-3}L\right)\\ \\ =2.2\times {10}^{-3}.\end{array}$

As there is two molecules of $HCl$, moles of added $HCl$ will be $4.4\times {10}^{-3}$. 

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{No.ofmolesBa{\left(OH\right)}_2\left(aq\right)+2HCl\left(aq\right)\to 2H_{2}O\left(l\right)+BaC{l}_2\left(aq\right)}\\ Initial3.75\times {10}^{-3}4.4\times {10}^{-3}-0\\ Change-3.75\times {10}^{-3}-3.75\times {10}^{-3}-+3.75\times {10}^{-3}\end{array}}\\ Final00.65\times {10}^{-3}-3.75\times {10}^{-3}\end{array}$

$\begin{array}{l}Excess\left[H_3{O}^{+}\right]=\frac{0.65\times {10}^{-3}moles}{35.00\times {10}^{-3}L}\\ \\ =0.0185M.\end{array}$

$\begin{array}{l}pH=-\log \left[H_3{O}^{+}\right]\\ \\ =-\log \left(0.0185\right)\\ \\ =\mathrm{1.7311.}\end{array}$

Therefore, the $pH$ of the solution has been calculated to be $\text{1}\text{.7311}$. 

Now we can plot the graph as shown in figure 1. 

$\begin{array}{cc}Volumeofaddedacid/mL& pH\\ 0& 13.1761\\ 5& 12.7127\\ 8.522& 7.0000\\ 10& 1.7311\end{array}$