Hello I need help to confirm 7 and 8. Then find the value for 9 7. ? 8. ? 9. ? B

Question

Hello I need help to confirm 7 and 8. Then find the value for 9 7. ? 8. ? 9. ? B. Molar Soluhility of Calcium Hydroxide in the Presence of a Common lon Trial 1 Trial 2 Trial 3 1. Volume of saturated Car(OH), with added CaCl, solution (ml) 2. Concentration of standardized HCI solution (mol/L) A Buret reading, initial (ml) Buret reading,jinal (mal) 5. Volume of HCI added (ml) & Moles of HCl added (mo 7. Moles of OH in saturated solution (mob2xD & IOH 1 equilibrium (mol/L) 9. Molar solubility of Ca(OH), with added 25.0 25.0 25.0 as4 10. Average molar solubility of CarOH), with added CaCl (mol/) Account for the different molar solubilities in Part A.11 and Part B.10 (Report Sheet)

Explanation / Answer

7 is correct  

8   the value of [OH-] = moles / volume of Ca(OH)2 added in L

[OH-] = 2 x 10^ -4 / ( 0.025) = 8 x 10^-3 M

[OH-] second coloumn = ( 1.5x10^-4) / ( 0.025) = 6 x 10^-3 M

hence [ OH-] = 2.5x10^-4 / 0.025 = 10^-2 M

9)   For Ca(OH)2 (s) <---> ca2+ (aq) + 2OH- (aq)

Solubility = S = [Ca2+] and [OH-] = 2S , but by neutralisation we found total [OH-] which is twice amount of S , hence S = 1/2 [OH-]

hence Molar solubility of Ca(OH)2 is   4 x 10^ -3 , 3 x 10^ -3 , 5 x 10^ -3

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