This question is for a laboratory experiment titled \"Determination of the Solub

Question

This question is for a laboratory experiment titled "Determination of the Solubility Product Constant of Calcium Hydroxide" where a saturated solution of Ca(OH)2 in 0.02523M NaOH was titrated with HCl.

Part 2 Data - Saturated Solution of Ca(OH)2 in 0.02523M NaOH:

Volume of Ca(OH)2/NaOH aliquot: 25.00mL

Concentration of standard HCl: 0.1342M

Indicator Used: Bromothymol Blue

Average volume of HCl to reach end point: 10.26mL

a. Calculate the TOTAL [OH-] in the saturated solution of Ca(OH)2 in sodium hydroxide for the solution assigned (Ca(OH)2 in 0.02523M NaOH).

b. Calculate the [OH-] that comes from the dissolution of Ca(OH)2. The total [OH-] (calculated above) is the sum of the [OH-] from the NaOH and the [OH-] from Ca(OH)2.

c. Calculate the solubility of Ca(OH)2 in the NaOH solution (in mol/L).

d. Calculate the experimental Ksp of Ca(OH)2 for the saturated solution of Ca(OH)2 in NaOH.

Please explain steps/show equations! Thank you.

Explanation / Answer

1) using the formula

M1V1 (HCL) = M2V2 (Ca(OH)2/NaOH)

0.1342M * 10.26mL = M2 * 25.00mL

M2 = 0.1342M * 10.26mL / 25.00mL = 0.05508 M

molarity of Ca(OH)2/NaOH aliquot = 0.05508 M

hence the concentration  the TOTAL [OH-] in the saturated solution of Ca(OH)2 in sodium hydroxide for the solution assigned (Ca(OH)2 in 0.02523M NaOH).=  0.05508 M   

2) AS the concentration of total [OH]- =  [OH-] from the NaOH + [OH-] from Ca(OH)2.

[OH-] from Ca(OH)2. =  concentration of total [OH-] - [OH-] from the NaOH

=  0.05508 M - 0.02523M = 0.02885 M

[OH-] from Ca(OH)2. =

0.02885 M

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