fernandapinkinsons fernandapinkinsons
  • 11-11-2016
  • Mathematics
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What volume of a 0.500 M HCl solution is needed to neutralize each: 14.0 ml of a 0.300 M NaOH and 19.0ml of a 0.200 M Ba(OH)2

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taskmasters
taskmasters taskmasters
  • 15-11-2016
First thing, we need to know the chemical equation for each neutralization reaction. Then, determine the amount in moles needed hydrochloric acid. We do as follows:

HCl + NaOH = NaCl + H2O

 0.300 M NaOH ( .014 L ) ( 1 mol  HCl / 1 mol NaOH ) ( 1 L / 0.500 M HCl ) = 0.0084 L HCl solution or 8.4 mL of 0.500 M HCl

2HCl + Ba(OH)2 = BaCl2 + 2H2O

0.200 M Ba(OH)2 (0.019 L ) ( 2 mol HCl / 1 mol Ba(OH)2 ) 
( 1 L / 0.500 M HCl ) = 0.0152 L HCl solution or 15.2 mL of 0.500 M HCl
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