Hard water contains alkaline earth cations such as Ca2+, which react with CO32- to form insoluble deposits of CaCO3. Will a precipitate of CaCO3 form if a 250 mL sample of hard water having [Ca2+] = 8.0 x 10^-4 M is treated with the following? (a) 0.10 mL of 2.0 x 10^-3 M Na2CO3 (b) 10 mg of solid Na2CO3
Verified step by step guidance
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Step 1: Calculate the moles of Ca^{2+} in the 250 mL sample of hard water using the concentration [Ca^{2+}] = 8.0 \times 10^{-4} \text{ M}.
Step 2: For part (a), calculate the moles of CO_3^{2-} added by using the volume (0.10 mL) and concentration (2.0 \times 10^{-3} \text{ M}) of Na_2CO_3 solution.
Step 3: For part (b), convert the mass of Na_2CO_3 (10 mg) to moles using its molar mass, and then determine the moles of CO_3^{2-} provided.
Step 4: Determine the reaction quotient Q_{sp} for the potential formation of CaCO_3 using the concentrations of Ca^{2+} and CO_3^{2-} in the solution.
Step 5: Compare Q_{sp} to the solubility product constant K_{sp} of CaCO_3. If Q_{sp} > K_{sp}, a precipitate will form; if Q_{sp} < K_{sp}, no precipitate will form.