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Since Calcuium hydroxide has a very strong dissassociation forevery .pdf

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Since Calcuium hydroxide has a very strong dissassociation forevery mole of calcium hydroxide is 2 moles of OH-. Ca(OH)2 ---> Ca2+ +2(OH)- 2*.03=.06 mol/L pOH = - log [OH-] pOH= -log .06 pOH= 1.22. Now pH + pOH = 14 So pH=14-1.22 pH= 12.78 Solution Since Calcuium hydroxide has a very strong dissassociation forevery mole of calcium hydroxide is 2 moles of OH-. Ca(OH)2 ---> Ca2+ +2(OH)- 2*.03=.06 mol/L pOH = - log [OH-] pOH= -log .06 pOH= 1.22. Now pH + pOH = 14 So pH=14-1.22 pH= 12.78.

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Since Calcuium hydroxide has a very strong dissassociation forevery mole of calcium hydroxide is 2 moles of OH-. Ca(OH)2 ---> Ca2+ +2(OH)- 2*.03=.06 mol/L pOH = - log [OH-] pOH= -log .06 pOH= 1.22. Now pH + pOH = 14 So pH=14-1.22 pH= 12.78 Solution Since Calcuium hydroxide has a very strong dissassociation forevery mole of calcium hydroxide is 2 moles of OH-. Ca(OH)2 ---> Ca2+ +2(OH)- 2*.03=.06 mol/L pOH = - log [OH-] pOH= -log .06 pOH= 1.22. Now pH + pOH = 14 So pH=14-1.22 pH= 12.78

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  • 1. Since Calcuium hydroxide has a very strong dissassociation forevery mole of calcium hydroxide is 2 moles of OH-. Ca(OH)2 ---> Ca2+ +2(OH)- 2*.03=.06 mol/L pOH = - log [OH-] pOH= -log .06 pOH= 1.22. Now pH + pOH = 14 So pH=14-1.22 pH= 12.78 Solution Since Calcuium hydroxide has a very strong dissassociation forevery mole of calcium hydroxide is 2 moles of OH-. Ca(OH)2 ---> Ca2+ +2(OH)- 2*.03=.06 mol/L pOH = - log [OH-] pOH= -log .06 pOH= 1.22. Now pH + pOH = 14 So pH=14-1.22 pH= 12.78