When 100 mL of 1.91 M K2CO3 and 565. mL of 1.31 M Ca(OH)2 are mixed a precipitate forms. Write the the balanced formula to help answer the two questions.1. What is the mass in grams of the precipitate?2. What is the concentration of the OH- ions in the final solution? Report answer to 3 sig figs If there are no longer there enter 0
Question
When 100 mL of 1.91 M K2CO3 and 565. mL of 1.31 M Ca(OH)2 are mixed a precipitate forms. Write the the balanced formula to help answer the two questions.1. What is the mass in grams of the precipitate?2. What is the concentration of the OH- ions in the final solution? Report answer to 3 sig figs If there are no longer there enter 0
Solution
First, let's write the balanced chemical equation for the reaction:
2 K2CO3 (aq) + 2 Ca(OH)2 (aq) → 2 CaCO3 (s) + 4 KOH (aq)
- To find the mass of the precipitate (CaCO3), we first need to determine the limiting reactant.
Moles of K2CO3 = Volume (L) x Molarity = 0.1 L x 1.91 mol/L = 0.191 mol Moles of Ca(OH)2 = Volume (L) x Molarity = 0.565 L x 1.31 mol/L = 0.740 mol
From the balanced equation, we see that the reaction ratio of K2CO3 to Ca(OH)2 is 1:1. Therefore, K2CO3 is the limiting reactant.
The moles of CaCO3 formed = moles of limiting reactant = 0.191 mol
The molar mass of CaCO3 = 40.08 g/mol (for Ca) + 12.01 g/mol (for C) + 3*16.00 g/mol (for O) = 100.09 g/mol
So, the mass of the precipitate = moles x molar mass = 0.191 mol x 100.09 g/mol = 19.1 g
- The concentration of OH- ions in the final solution is determined from the KOH produced. From the balanced equation, we see that 2 moles of K2CO3 produce 4 moles of KOH. Therefore, 0.191 mol of K2CO3 will produce 0.191 mol x 2 = 0.382 mol of KOH.
Since each KOH has one OH- ion, the moles of OH- = 0.382 mol.
The total volume of the solution = volume of K2CO3 + volume of Ca(OH)2 = 0.1 L + 0.565 L = 0.665 L
So, the concentration of OH- ions = moles/volume = 0.382 mol / 0.665 L = 0.574 M
Therefore, the mass of the precipitate is 19.1 g and the concentration of OH- ions in the final solution is 0.574 M.
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