Chapter 8, Problem 188CP

Interpretation Introduction

Interpretation:

The pH of the solution formed by mixing the given solutions needs to be determined.

Concept introduction:

The pH of a solution may be stated as it is negative logarithm of concentration of H+ ion. Mathematically it can be written as:

  $pH=-log\left[H^{+}\right]$

Answer to Problem 188CP

The pH of solution is 7.46.

Explanation of Solution

Number of moles of H₂SO₄ in 50.0 mL of 0.100 M H₂SO₄ is calculated as below −

  $\begin{array}{l}Moleof{H}_{2}S{O}_4=Concentration\left(Molarity\right)of{H}_{2}S{O}_4\times Volumeof{H}_{2}S{O}_{4}solutioninLiter\hfill \\ Moleof{H}_{2}S{O}_4=\left(0.1M\right)\times \left(0.05L\right)\hfill \\ Moleof{H}_{2}S{O}_4=\left(0.1mol/L\right)\times \left(0.05L\right)\hfill \\ Moleof{H}_{2}S{O}_4=\left(0.005mol\right)\hfill \end{array}$

So, 50.0 mL of 0.100 M H₂SO₄ contains (0.005 mol ) of H₂SO₄ .

(II) Number of moles of HOCl in 30.0 mL of 0.100 M HOCl is calculated as below −

  $\begin{array}{l}MoleofHOCl=Concentration\left(Molarity\right)ofHOCl\times VolumeofHOClsolutioninLiter\hfill \\ MoleofHOCl=\left(0.1M\right)\times \left(0.03L\right)\hfill \\ MoleofHOCl=\left(0.1mol/L\right)\times \left(0.03L\right)\hfill \\ MoleofHOCl=\left(0.003mol\right)\hfill \end{array}$

So, 30.0 mL of 0.100 M HOCl contains (0.003 mol ) of HOCl .

(III) Number of moles of NaOH in 25.0 mL of 0.200 M NaOH is calculated as below −

  $\begin{array}{l}MoleofNaOH=Concentration\left(Molarity\right)ofNaOH\times VolumeofNaOHsolutioninLiter\hfill \\ MoleofNaOH=\left(0.2M\right)\times \left(0.025L\right)\hfill \\ MoleofNaOH=\left(0.2mol/L\right)\times \left(0.025L\right)\hfill \\ MoleofNaOH=\left(0.005mol\right)\hfill \end{array}$

So, 25.0 mL of 0.200 M NaOH contains ( 0.005 mol ) of NaOH .

(IV) Number of moles of Ba(OH)₂ in 25.0 mL of 0.100 M Ba(OH)₂ is calculated as below -

  $\begin{array}{l}MoleofBa{\left(OH\right)}_2=Concentration\left(Molarity\right)ofBa{\left(OH\right)}_2\times VolumeofBa{\left(OH\right)}_{2}solutioninLiter\hfill \\ MoleofBa{\left(OH\right)}_2=\left(0.1M\right)\times \left(0.025L\right)\hfill \\ MoleofBa{\left(OH\right)}_2=\left(0.1mol/L\right)\times \left(0.025L\right)\hfill \\ MoleofBa{\left(OH\right)}_2=\left(0.0025mol\right)\hfill \end{array}$

(V ) Number of moles of KOH in 10.0 mL of 0.150 M KOH is calculated as below −

  $\begin{array}{l}MoleofKOH=Concentration\left(Molarity\right)ofKOH\times VolumeofKOHsolutioninLiter\hfill \\ MoleofKOH=\left(0.15M\right)\times \left(0.01L\right)\hfill \\ MoleofKOH=\left(0.15mol/L\right)\times \left(0.01L\right)\hfill \\ MoleofKOH=\left(0.0015mol\right)\hfill \\ \hfill \end{array}$

So, 10.0 mL of 0.150 M KOH contains (0.0015 mol ) of KOH .

Now,

  $\begin{array}{l}Totalmoleof{H}^{+}ion=\hfill \\ Moleof{H}^{+}ionin50.0mLof0.100M{H}_{2}S{O}_4+Moleof{H}^{+}ionin30.0mLof0.100MHOCl\hfill \\ =\left(2\times 0.005mol\right)+\left(0.003mol\right)\hfill \\ =>Totalmoleof{H}^{+}ion=\left(0.01mol\right)+\left(0.003mol\right)\hfill \\ =>Totalmoleof{H}^{+}ion=\left(0.013mol\right)\hfill \\ \hfill \end{array}$

  $\begin{array}{l} TotalmoleofO{H}^{-}ion=\hfill \\ MoleofO{H}^{-}ionin25.0mLof0.200MNaOH+MoleofO{H}^{-}ionin25.0mLof0.100MBa{\left(OH\right)}_2\hfill \\ +10.0mLof0.150MKOH=\left(0.005mol\right)+\left(2\times 0.0025mol\right)+\left(0.0015mol\right)\hfill \\ =>TotalmoleofO{H}^{-}ion=\left(0.005mol\right)+\left(0.005mol\right)+\left(0.0015mol\right)\hfill \\ =>TotalmoleofO{H}^{-}ion=\left(0.0115mol\right)\hfill \end{array}$

Left moles of OH- = 0.0115 -0.01=0.0015mol

  $HOCl+O{H}^{-}\to OC{l}^{-}+H_3{O}^{+}$

	HOCl	OH-	OCl-	H3O+
I	0.003	0.0015
C	-0.0015	-0.0015	0.0015	0.0015
E	0.0015	0	0.0015	0.0015

  $\begin{array}{l}pH=p{k}_a+\log \frac{\left[OC{l}^{-}\right]}{\left[HOCl\right]}\\ Ka\text{ of HOCl}=3.5\times {10}^{-8}\\ pKa=-\log Ka=-\log 3.5\times {10}^{-8}\\ =7.46\\ pH=p{k}_a+\log \frac{\left[0.0015\right]}{\left[0.0015\right]}\\ =7.46+0\\ =7.46\end{array}$

Conclusion

Thus, the pH of solution is 7.46.