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valentina_108 [34]
3 years ago
9

How concentrated can lye solution be before its ph rises above calcium hydroxide saturated water?

Chemistry
1 answer:
Flauer [41]3 years ago
5 0
The solubility of calcium hydroxide (Ca(OH)2) in water at 20^{o}C is 0.02335 <span>mol/L. When a Ca(OH)2 solution has a concentration of 0.02335 mol/L at the said temperature, a saturated solution is formed. 

Upon dissolution and dissociation of 1 mole Ca(OH)2, 1 mole of calcium ion (</span>Ca^{+2}) and 2 moles of hydroxide ions (OH^{-1}) are formed. The concentration of the hydroxide ions determine the pH of the solution. This same concentration would be needed in calculating the concentration of lye solution needed. 

1. To calculate the pH of the solution, a few approaches are possible. Here, the pOH of the solution is calculated first using the molar concentration of OH^{-1} ions. Because pH + pOH = 14, the pH of the solution can then be calculated by subtracting pOH from 14. 

Because 2 moles of OH^{-1} ions are formed per mole of Ca(OH)2 dissolved, the pOH of a saturated Ca(OH)2 solution at 20^{o}C is,  

*Concentration (C_{OH_{1}}) of OH^{-1} ions,
             C_{OH_{1}} = 0.02335 mol Ca(OH)2/L  * 2 mol OH^{-1}/ 1 mol Ca(OH)2
             C_{OH_{1}} = 0.0467 mol OH^{-1}/L 

*pOH of saturated Ca(OH)2 solution, 
                                    pOH = -log [C_{OH_{1}}]
                                    pOH = -log [0.0467 mol OH^{-1}/L ]             
                                    pOH = 1.33        

*pH of saturated Ca(OH)2 solution,
                                    pH = 14 - 1.33
                                    pH = 12.67

2. To find the concentration of lye (NaOH) solution that would give the same pH as a saturated Ca(OH)2 solution at the same temperature, the pOH of the saturated Ca(OH)2 solution will be used. 

For every mole of NaOH that dissolves and dissociates, 1 mole of OH^{-1} ions is formed. Thus, 

*Concentration (
C_{OH_{2}}) of OH^{-1} ions,
             pOH = -log [C_{OH_{2}}]
               1.33 = -log [C_{OH_{2}} ]       
             C_{OH_{2}} = 10^-1.33   
             C_{OH_{2}} = 0.0467 mol OH^{-1}/ L

Because 1 mole of NaOH produces only 1 mole of OH^{-1} ions, the concentration of the OH^{-1} ions should then be equal to the concentration of the lye (NaOH) solution. Thus, 0.0467 mol/L of lye would give the same pH as that of 0.0233 mol/L (saturated) Ca(OH)2 solution at 20^{o}C. The pH of both solutions is 12.67.  
      
                                               
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Use proportions:

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=> x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3

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Formula Used,

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