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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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paadms214 | Student, Grade 11 | (Level 3) eNoter

Posted February 12, 2012 at 6:15 AM via web

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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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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 12, 2012 at 12:35 PM (Answer #1)

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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.

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