Compute the ph when 5.86 g of naclo is added hclo

Reference no: EM13334816

Calculate the pH when 5.86 g of NaCLO is added to 454.1 mL of .5M HCLO (ignoring any changes in volume). Given: Ka value for HClO: 3.0 X 10^-8. This is what I did: 1) calculated molarity of HClO: .50/.4541=.174 2) calculated molarity of NaClO: (5.86g/74.45 g/mol)/(.4541L=1.10) 3) Set up ICE chart and calculated H+(x). x=(3E-8)(1.10)/(.174) = 1.90E-7 4) took negative log of 1.90E-7 to get a pH of 6.72.

Reference no: EM13334816

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Compute the ph when 5.86 g of naclo is added hclo : Calculated molarity of HClO: .50/.4541=.174 2) calculated molarity of NaClO: (5.86g/74.45 g/mol)/(.4541L=1.10) 3) Set up ICE chart and calculated H+(x). x=(3E-8)(1.10)/(.174) = 1.90E-7 4) took negative log of 1.90E-7 to get a pH of 6.72.

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