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What volume of 0.585 M Ca(OH)2 would be needed to neutralize 15.8 L of 1.51 M HCl?
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One molecule of HCl dissociates to give one ion of H+ and one in of Cl-. Ca(OH)2 on the other hand dissociates to give one ion of Ca2+ and 2 ions of (OH)-.
During the neutralization of Ca(OH)2 and HCl, the H+ and (OH)- ions combine to form one molecule of water.
In a 15.8 L solution of 1.51 M HCl there are 1.51*15.8 = 23.858 moles of HCl. This gives 23.858 moles of H+ ions. Let the volume of 0.585M Ca(OH)2 be V. The number of moles of Ca(OH)2 in this should be half the number of moles of HCl.
=> V*0.585 = 23.858/2
=> V = 20.39
20.39 L of 0.585 M Ca(OH)2 is required to neutralize 15.8 L of 1.51 M HCl.
Posted by justaguide on February 12, 2012 at 12:35 PM (Answer #1)
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