91Ó°ÊÓ

First, let's convert the given volumes and concentrations to moles for each acid and base: - Moles of H2SO4: \((50.0\,mL)(0.100\, M) = 5.00\,mmol\) - Moles of HOCL: \((30.0\,mL)(0.100\, M) = 3.00\,mmol\) - Moles of NaOH: \((25.0\,mL)(0.200\, M) = 5.00\,mmol\) - Moles of Ba(OH)2: \((25.0\,mL)(0.100\, M) = 2.50\,mmol\) - Moles of KOH: \((10.0\,mL)(0.150\, M) = 1.50\,mmol\)

Water is a solvent in which all acids and bases are mixed and form the solution. The final volume of the solution can be calculated by adding the volumes of individual components: - Total Volume: \(50.0 + 30.0 + 25.0 + 25.0 + 10.0 = 140.0\,mL\)

We will carry out the reactions between different acids and bases. There are three acids, \(\ce{H2SO4}\), \(\ce{HOCl}\), and \(\ce{H2O}\) (since it's an aqueous solution), and three bases, \(\ce{NaOH}\), \(\ce{Ba(OH)2}\), and \(\ce{KOH}\). The neutralization reactions are as follows: 1. Reaction between \(\ce{H2SO4}\) and strong bases \(\ce{NaOH}\), \(\ce{Ba(OH)2}\), and \(\ce{KOH}\): Since \(\ce{H2SO4}\) is a strong diprotic acid, it will react with all three strong bases until completely neutralized. - \(\ce{H2SO4}\) reacts with \(\ce{NaOH}\): \(5.00\, mmol \, \ce{H2SO4} + 5.00\, mmol \, \ce{NaOH} \rightarrow 0\, mmol \, \ce{H2SO4}\) and \(0\, mmol \, \ce{NaOH}\) - \(\ce{H2SO4}\) is completely neutralized; there is no \(\ce{H2SO4}\) left. 2. Reaction between \(\ce{HOCL}\) and strong bases \(\ce{Ba(OH)2}\) and \(\ce{KOH}\): Since \(\ce{HOCL}\) is a weak monoprotic acid, it will react with the remaining strong bases. - \(\ce{HOCL}\) reacts with \(\ce{Ba(OH)2}\): \(3.00\, mmol \, \ce{HOCL} + 2.50\, mmol \, \ce{Ba(OH)2} \rightarrow 0.50\, mmol \, \ce{HOCL}\) and \(0\, mmol \, \ce{Ba(OH)2}\) - \(\ce{HOCL}\) reacts with \(\ce{KOH}\): \(0.50\, mmol \, \ce{HOCL} + 1.50\, mmol \, \ce{KOH} \rightarrow 0\, mmol \, \ce{HOCL}\) and \(1.00\, mmol \, \ce{KOH}\) - \(\ce{HOCL}\) is completely neutralized; there is no \(\ce{HOCL}\) left. 3. Reaction between \(\ce{H2O}\) and strong base \(\ce{KOH}\): - No reaction will happen, as water will not react with strong bases under these conditions.

After the reactions, we have the following concentrations of the remaining species: - \(\ce{OH^-}\) concentration from \(\ce{KOH}\): \(\dfrac{1.00\,mmol}{140.0\,mL} = 0.00714\,M\)

Now we can calculate the pH of the solution. First, let's calculate the pOH: - pOH = \(-\log_{10}([OH^-]) = -\log_{10}(0.00714) = 2.15\) Since pH and pOH are related by the equation: \(pH + pOH = 14\), we can find the pH of the solution: - pH = \(14 - 2.15 = 11.85\) The pH of the solution is 11.85.