If 20.0 cm^3 of 0.20m Sodium hydroxide (NaOH) solution is mixed with 1.00 dm^3 of 0.10M calcium chloride (CaCl2) solution, determine whether a precipitate of Ca(OH)2 will form or not ? Show working. [Note: Ksp (Ca(OH)2) = 6.4×10^-6]
Answers
Answer:
NaOH
NaOH[OH] = 2M
NaOH[OH] = 2M//////////////////////////
NaOH[OH] = 2M//////////////////////////Ca(OH) = 74 g/mol
NaOH[OH] = 2M//////////////////////////Ca(OH) = 74 g/molCa(OH)2 solubility 1.73 g/L
NaOH[OH] = 2M//////////////////////////Ca(OH) = 74 g/molCa(OH)2 solubility 1.73 g/L1.73 / 74 = 0.023mol
NaOH[OH] = 2M//////////////////////////Ca(OH) = 74 g/molCa(OH)2 solubility 1.73 g/L1.73 / 74 = 0.023molM = 0.023 mol/L
NaOH[OH] = 2M//////////////////////////Ca(OH) = 74 g/molCa(OH)2 solubility 1.73 g/L1.73 / 74 = 0.023molM = 0.023 mol/L[OH] = 2 x 0.023 M = 0.046M
NaOH[OH] = 2M//////////////////////////Ca(OH) = 74 g/molCa(OH)2 solubility 1.73 g/L1.73 / 74 = 0.023molM = 0.023 mol/L[OH] = 2 x 0.023 M = 0.046M///////////////////////////////////////
NaOH[OH] = 2M//////////////////////////Ca(OH) = 74 g/molCa(OH)2 solubility 1.73 g/L1.73 / 74 = 0.023molM = 0.023 mol/L[OH] = 2 x 0.023 M = 0.046M///////////////////////////////////////the [OH] is more higher in the solution NaOH
NaOH[OH] = 2M//////////////////////////Ca(OH) = 74 g/molCa(OH)2 solubility 1.73 g/L1.73 / 74 = 0.023molM = 0.023 mol/L[OH] = 2 x 0.023 M = 0.046M///////////////////////////////////////the [OH] is more higher in the solution NaOHCa(OH)2 is a trong base but not too soluble in water. So it can not male 1M solution