A solution is made by adding $1.000 \mathrm{~g} \mathrm{Ca}(\mathrm{OH})_{2}(s), 100.0 \mathrm{~mL}$ of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\), and enough water to make a final volume of \(350.0 \mathrm{~mL}\). Assuming that all of the solid dissolves, what is the \(\mathrm{pH}\) of the final solution?

Short Answer

Expert verified
The pH of the final solution is approximately 12.16.

Step by step solution

01

1. Calculate the moles of Ca(OH)2 and H2SO4

First, find the moles of Ca(OH)2 present in the solution by using its mass and molar mass: Moles of Ca(OH)2 = mass of Ca(OH)2 / molar mass of Ca(OH)2 Moles of Ca(OH)2 = 1.000 g / (40.08 + 2 * (15.999 + 1.008)) g/mol Moles of Ca(OH)2 ≈ 0.0125 mol Next, find the moles of H2SO4 present in the solution by using its concentration and volume: Moles of H2SO4 = concentration of H2SO4 * volume of H2SO4 Moles of H2SO4 = 0.10 M * 100.0 mL * (1 L / 1000 mL) Moles of H2SO4 = 0.010 mol
02

2. Determine the moles of OH- ions remaining after the reaction

The balanced chemical equation for the reaction between Ca(OH)2 and H2SO4 is: Ca(OH)2(s) + H2SO4(aq) → CaSO4(aq) + 2 H2O(l) From the balanced equation, know that 1 mole of Ca(OH)2 reacts with 1 mole of H2SO4. Since there's an excess of Ca(OH)2, some of it will remain unreacted at the end. Therefore, the final solution will be basic. Calculate the moles of Ca(OH)2 remaining after the reaction: Moles of Ca(OH)2 remaining = moles of Ca(OH)2 - moles of H2SO4 Moles of Ca(OH)2 remaining ≈ 0.0125 - 0.0100 Moles of Ca(OH)2 remaining ≈ 0.0025 mol Now, calculate the total moles of OH- ions in the remaining Ca(OH)2: Moles of OH- ions = 0.0025 mol Ca(OH)2 * 2 moles OH- per 1 mol Ca(OH)2 Moles of OH- ions = 0.0050 mol
03

3. Calculate the concentration of OH- ions in the final solution

We need the volume of the final solution to find the concentration of OH- ions. Final Volume = 100.0 mL (H2SO4) + 350.0 mL (total volume) - 100.0 mL (initial volume) = 350.0 mL Now, calculate the concentration of OH- ions: Concentration of OH- ions = moles of OH- ions / volume of final solution Concentration of OH- ions = 0.0050 mol / 350.0 mL * (1000 mL / 1 L) Concentration of OH- ions = 0.0143 M
04

4. Calculate the pOH and pH of the final solution

Use the concentration of OH- ions to find the pOH: pOH = -log10(concentration of OH- ions) pOH = -log10(0.0143) pOH ≈ 1.84 Lastly, use the relationship between pOH and pH to find the pH of the final solution: pH = 14 - pOH pH = 14 - 1.84 pH ≈ 12.16 So the pH of the final solution is approximately 12.16.

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Most popular questions from this chapter

The active ingredient in aspirin is acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right),\) a monoprotic acid with \(K_{a}=3.3 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) What is the \(\mathrm{pH}\) of a solution obtained by dissolving one regular aspirin tablet, containing \(100 \mathrm{mg}\) of acetylsalicylic acid, in $200 \mathrm{~mL}$ of water?

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