Question

Consider a solution formed by mixing \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}, 30.0 \mathrm{~mL}\) of \(0.100 \mathrm{MOCl}, 25.0 \mathrm{~mL}\) of \(0.200 \mathrm{MaOH}\), \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\), and \(10.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{KOH}\). Calculate the \(\mathrm{pH}\) of this solution.

Short answer

The pH of the final solution is approximately 11.85.

Step-by-step solution

Calculate the moles of each component

We need to determine the number of moles of each compound in the solution. To do this, we will use the formula: moles = molarity × volume For H2SO4: moles = 0.100 M × 50.0 mL = 5.00 mmol For OCl-: moles = 0.100 M × 30.0 mL = 3.00 mmol For NaOH: moles = 0.200 M × 25.0 mL = 5.00 mmol For Ba(OH)2: moles = 0.100 M × 25.0 mL = 2.50 mmol (keep in mind each mole of Ba(OH)2 has 2 moles of OH-, so it will be 5.00 mmol of OH-) For KOH: moles = 0.150 M × 10.0 mL = 1.50 mmol

Perform stoichiometric calculations

First, consider H2SO4 and NaOH reaction: H2SO4 + 2NaOH → Na2SO4 + 2H2O H2SO4 moles: 5.00 mmol - 5.00 mmol = 0 mmol (consumed) NaOH moles: 5.00 mmol - 5.00 mmol = 0 mmol (consumed) Now consider the remaining H2SO4 and Ba(OH)2 reaction: H2SO4 + Ba(OH)2 → BaSO4 + 2H2O H2SO4 moles: 0 mmol (no H2SO4 left, the reaction does not occur) Now consider the reaction between OCl- and Ba(OH)2: OCl- + Ba(OH)2 → Ba(ClO)2 + 2H2O OCl- moles: 3.00 mmol - 2.50 mmol = 0.50 mmol (remaining) Ba(OH)2 moles: 2.50 mmol - 2.50 mmol = 0 mmol (consumed) Now consider the reaction between OCl- and KOH: OCl- + KOH → KClO + H2O OCl- moles: 0.5 mmol - 0.5 mmol = 0 mmol (consumed) KOH moles: 1.50 mmol - 0.5 mmol = 1.00 mmol (remaining)

Find the concentration of excess base

To find out the final concentration of the remaining KOH, we will use the formula: concentration = moles / final volume Final volume = (50.0 mL + 30.0 mL + 25.0 mL + 25.0 mL + 10.0 mL) = 140.0 mL Final concentration of KOH = 1.00 mmol / 140.0 mL = 0.00714 M

Calculate the pH of the solution

Now we can calculate the pOH of our solution using remaining KOH concentration, and then, finally, find the pH. pOH = -log10([OH-]) = -log10(0.00714) ≈ 2.15 pH = 14 - pOH = 14 - 2.15 = 11.85 In conclusion, the pH of the final solution is 11.85.

Key concepts

Solution Stoichiometry

Solution stoichiometry involves the quantitative aspects of chemical solutions, allowing chemists to predict the outcomes of reactions in solution. In practical terms, this involves calculating the number of moles of reactants and products using molarity (moles of solute per liter of solution) and volume.

To understand stoichiometry in solutions, it is beneficial to become familiar with the concept of the mole, as it relates to the amount of a substance. When combined with balanced chemical equations, stoichiometry allows us to determine how much of each reactant is needed and how much of each product will be formed. For instance, in the initial exercise, the stoichiometry of reactions between H₂SO₄, NaOH, Ba(OH)₂, and KOH is used to calculate the resultant moles after the reactions have taken place.

Acid-Base Titration

Acid-base titration is a process used to determine the concentration of an acid or a base by reacting it with a base or acid of known concentration. It involves the careful addition of one solution to another until a neutralization reaction is achieved, indicating equivalent molar amounts of H⁺ and OH^- have combined to form water.

In the context of our exercise, although a titration is not performed, the principles apply. We're essentially mixing known volumes and molarities of acids and bases to reach a state where we can determine the excess reactants. Titrations typically require an indicator to show when equivalent amounts of acid and base have reacted; however, in solution stoichiometry calculations like ours, we use the balanced equations and stoichiometric calculations to identify the equivalence point.

Molarity and Volume Relationship

The relationship between molarity and volume is integral to calculating the quantities of reactants in solution chemistry. Molarity, defined as moles of a solute per liter of solution (mol/L), can be adjusted through dilution or concentration by changing the volume of the solution while keeping the amount of solute constant.

For instance, as seen in the given solution, the moles of each component in the mixture are calculated by multiplying the molarity by the volume in liters. This relationship is pivotal when combining solutions of different reactants, as it allows us to calculate the final concentration of the remaining species after the reactions have taken place, such as the remaining KOH in our exercise.

pOH and pH Calculations

pOH and pH are measures of the acidity and basicity of a solution, respectively. The pH scale typically ranges from 0 to 14, with lower values being more acidic and higher values more basic. The pOH scale is related to pH and represents the concentration of hydroxide ions (OH^-). The relationship between pOH and pH is given by the equation pH + pOH = 14.

To calculate pH, we often first find the pOH by taking the negative logarithm of the hydroxide ion concentration of the solution. We then use the aforementioned relationship to find the pH. In the provided exercise, we calculated the remaining concentration of KOH (a strong base) after the reactions and used it to find the pOH, subsequently determining the pH of the solution to be 11.85, indicating a basic solution.