Question

Consider $1.0\mathrm{L}$ of an aqueous solution that contains $0.10$ $M$ sulfuric acid to which 0.30 mole of barium nitrate is added. Assuming no change in volume of the solution, determine the $\mathrm{pH},$ the concentration of barium ions in the final solution, and the mass of solid formed.

Short answer

The pH of the final solution is 0.70, the concentration of barium ions in the final solution is $0.20\mathrm{M}$, and the mass of solid barium sulfate formed is $23.34\mathrm{g}$.

Step-by-step solution

Write the balanced reaction equation

The balanced reaction equation between sulfuric acid and barium nitrate is: ${\mathrm{H}}_2{\mathrm{SO}}_4+\mathrm{Ba}({\mathrm{NO}}_3{)}_2\to {\mathrm{BaSO}}_4(\mathrm{s})+2{\mathrm{HNO}}_3$ This tells us that 1 mole of sulfuric acid reacts with 1 mole of barium nitrate to produce 1 mole of barium sulfate and 2 moles of nitric acid.

Calculate the moles of sulfuric acid in the initial solution

First, let's find the moles of sulfuric acid (${\mathrm{H}}_2{\mathrm{SO}}_4$) in the 1.0 L solution: Moles = Molarity × Volume = $0.10\mathrm{M}\times 1.0\mathrm{L}=0.10\mathrm{mol}$

Determine the limiting reactant and calculate the moles of products

We are given 0.30 mole of barium nitrate ($\mathrm{Ba}({\mathrm{NO}}_3{)}_2$), and calculated 0.10 mole of sulfuric acid. The balanced reaction equation tells us that 1 mole of sulfuric acid reacts with 1 mole of barium nitrate, so sulfuric acid is the limiting reactant. The moles of barium sulfate (${\mathrm{BaSO}}_4$) that are formed will be equal to the moles of limiting reactant (sulfuric acid): Moles of ${\mathrm{BaSO}}_4$ = 0.10 mol The moles of nitric acid (${\mathrm{HNO}}_3$) produced will be double the moles of limiting reactant: Moles of ${\mathrm{HNO}}_3$ = 2 × 0.10 mol = 0.20 mol

Calculate the final concentrations of hydronium ions (${\mathrm{H}}^{+}$) and barium ions (${\mathrm{Ba}}^{2+}$)

Since there is no change in the volume of the solution, the concentration of hydronium ions from nitric acid will be equal to the concentration of nitric acid (the acidity of the solution is now only due to nitric acid and the sulfuric acid has reacted completely): $$[{\mathrm{H}}^{+}]=\frac{0.20\mathrm{mol}}{1.0\mathrm{L}}=0.20\mathrm{M}$$ The concentration of barium ions in the final solution will be based on the initial amount of barium nitrate added and subtracting the moles that reacted with sulfuric acid: $$[{\mathrm{Ba}}^{2+}]=\frac{0.30\mathrm{mol}-0.10\mathrm{mol}}{1.0\mathrm{L}}=0.20\mathrm{M}$$

Calculate the pH of the final solution

Now we can calculate the pH of the solution using the concentration of hydronium ions: $$\mathrm{pH}=-\log ([{\mathrm{H}}^{+}])=-\log (0.20\mathrm{M})=0.70$$

Calculate the mass of solid barium sulfate formed

To find the mass of solid barium sulfate formed, we multiply the moles of barium sulfate by its molar mass: Mass = Moles × Molar mass = $0.10\mathrm{mol}\times 233.4\frac{\mathrm{g}}{\mathrm{mol}}=23.34\mathrm{g}$ #Conclusion#: The pH of the final solution is 0.70, the concentration of barium ions in the final solution is 0.20 M, and the mass of solid barium sulfate formed is 23.34 g.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is completely consumed first, stopping the reaction from continuing. It determines the maximum amount of product that can be formed.

In the given exercise, we are dealing with a reaction between sulfuric acid ( H_2SO_4 ) and barium nitrate ( Ba(NO_3)_2 ). The balanced reaction is:

• ext{H_2SO_4 + Ba(NO_3)_2 ightarrow BaSO_4(s) + 2HNO_3}

Here's how we find the limiting reactant:

• Calculate the moles of each reactant. We have 0.10 mole of H_2SO_4 and 0.30 mole of Ba(NO_3)_2 .
• According to the balanced equation, 1 mole of H_2SO_4 reacts with 1 mole of Ba(NO_3)_2 .
• Since we only have 0.10 mole of H_2SO_4 , it will be used up first, making it the limiting reactant.

Recognizing this helps us determine how much barium sulfate ( BaSO_4 ) can form and directs us to the next calculations.

pH Calculation

To find the pH of a solution, we need to calculate the concentration of H^+ ions. The pH tells us how acidic or basic a solution is.

Here's how the pH calculation unfolds for this exercise:

• Once the sulfuric acid reacts completely, it turns into nitric acid (HNO_3).
• We already established that 0.10 mole of H_2SO_4 produces 0.20 mole of HNO_3 because 1 mole of sulfuric acid forms 2 moles of nitric acid.
• Since volume doesn't change (1.0 L), the nitric acid concentration is 0.20 M.

Finally, we compute the pH using the formula: $pH=-\log ([H^{+}])=-\log (0.20)=0.70.$This shows that the resulting solution is quite acidic.

Precipitation Reactions

Precipitation reactions result in the formation of an insoluble solid from two soluble reactants in a solution.

For this exercise, the precipitation process involves:

• The reaction between H_2SO_4 and Ba(NO_3)_2 producing barium sulfate (BaSO_4) and nitric acid.
• Barium sulfate, the product of the reaction, precipitates as a solid because it is not soluble in water.
• The mass of BaSO_4 can be calculated using its molar mass (233.4 g/mol):

Mass of BaSO_4 is given by:
$Mass=\text{Moles}\times \text{Molar Mass}=0.10\times 233.4=23.34\text{ g}$Understanding precipitation is important for predicting which substances will remain dissolved and which will form solids.