solubility product of Ca(OH)2at 25'C is 4.42*10^-5. A 500ml of saturatee solution of it is mixed with an equal volume of 0.4M NaOH .How muchb Ca(OH)2 PREcipited
the solubility product constant (Ksp) of ca(oh)2 at 25 C is 4.42*10^-5. a 500 ml of saturated solution of ca(oh)2 is mixed with equal volume of .4M naoh. what mass of ca(oh)2 is precipitated?
Very hard question
0.754
To determine how much Ca(OH)2 is precipitated when 500 mL of a saturated solution of Ca(OH)2 is mixed with an equal volume of 0.4 M NaOH, we need to consider the solubility product and the chemical equation of the reaction between Ca(OH)2 and NaOH.
First, let's write the balanced chemical equation for the reaction:
Ca(OH)2 (saturated solution) + 2NaOH (aqueous) → Ca(OH)2 (precipitate) + 2NaOH (aqueous)
From the balanced equation, we can see that the stoichiometric ratio between Ca(OH)2 and NaOH is 1:2. This means that for every 1 mole of Ca(OH)2, 2 moles of NaOH are required to precipitate all the Ca(OH)2.
Now, let's calculate the number of moles of NaOH required for precipitating all the Ca(OH)2 in a 500 mL saturated solution:
Number of moles of NaOH = molarity × volume
= 0.4 M × 0.5 L (since we have equal volumes of saturated solution and NaOH)
= 0.2 moles
Since the stoichiometric ratio between Ca(OH)2 and NaOH is 1:2, we need twice the number of moles of Ca(OH)2 compared to NaOH. Therefore:
Number of moles of Ca(OH)2 = 2 × number of moles of NaOH
= 2 × 0.2 moles
= 0.4 moles
Finally, let's calculate the mass of Ca(OH)2 precipitated using its molar mass:
Mass of Ca(OH)2 = number of moles × molar mass
= 0.4 moles × (40.08 g/mol + 2 × 16.00 g/mol)
= 0.4 moles × 74.08 g/mol
= 29.63 g
Therefore, approximately 29.63 grams of Ca(OH)2 will be precipitated when the 500 mL saturated solution is mixed with an equal volume of 0.4 M NaOH.