5. Experimental Procedure, Part A.6 versus Part B.2. Would you expect more or le
ID: 1001575 • Letter: 5
Question
5. Experimental Procedure, Part A.6 versus Part B.2. Would you expect more or less standard 0.05 M HCI to be used to reach the methyl orange endpoint in Part B.2? Explain. 6. a. A determination of the molar solubility and the K calcium hydroxide was completed according to the experimen- tal procedure. The following data were collected for Trial 1. (See Report Sheet.) Complete the table for the deter- minations. Record the calculated values to the correct number of significant figures. A. Molar Solubility and Solubility Product of Calcium Hydroxide Calculation Zone 1. Volume of saturated Ca(OH) solution (mL) 2. Molar concentration of standard -2500 Part A.6 HCl solution (mol/L) 3. Buret reading, initial (mL) 4. Buret reading, final (mL) 5. Volume of HCI added (mL) 6. Moles of HCI added (mol) 0.0480 Part A. 7 1.20- 13.90 Part A.8 Show calculation. 5.01010 7. Moles of OH in saturated solution (mol) Show calculation. Part A.9 8. [OH1, equilibrium (mol/L) 9. ICa2*], equilibrium (mo/L) 10. Molar solubility of Ca(OH)2 (mol/L) .51 x12 Part A.10 .51 xlo2 Show calculation. -3 19x10 19X10 Show calculation. 1a x1o-3 Part A.11 Show calculation. 11. Kp of Ca(OH) Show calculation.Explanation / Answer
1) Volume of sat. Ca(OH)2 solution (mL) = 25.00
2) Vol of HCl added (mL) = 12.2
3) conc of HCl (mol/L) =0.0480
4) moles of HCl added (n) = N*V = 0.0480*12.2 = 0.0586
5) moles of [OH]- present at equlibrium is equal to [H]+ conc = 0.0586
6) Ca2+ ions concentration = 0.007262 = 7.262 *10-3
NCa(OH)2 * VCa(OH)2 = N HCl * VHCl
NCa(OH)2 = 0.0480*12.2/ 25
= 0.5856/25 = 0.023424 = 2.342 *10-2
Moles of Ca(OH)2 N*V =2n
n= 0.02324* 25/2 = 0.2905
moles of Ca2+ = 0.2905/40 = 0.007262 =7.262 * 10-3 moles
7) Ksp = [ Ca2+] * 2[OH]-
= s * (2s)2
= (7.262* 10-3 ) * (2 * 5.86 *10-4)2
= 99.75 * 10-10 (mol/L)3
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