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10.94 mL of KOH solution is required to titrate 0.5361 g of potassium hydrogen phthalate (KHP;...

10.94 mL of KOH solution is required to titrate 0.5361 g of potassium hydrogen phthalate (KHP; FW=204.224) to the Phenolphthalein endpoint. If 27.96 mL of this solution is required to titrate 96.34 mL of HCl to the Phenolphthalein endpoint, what is the concentration of the HCl solution?

__________ M

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Answer #1

moles of KOH = moles of KHP

moles of KHP = 0.5361 / 204.224

                      = 2.625 x 10^-3 mol

moles of KOH = 2.625 x 10^-3 mol

Molarity of KOH = 2.625 x 10^-3 / 0.01094

                         = 0.240 M

KOH +   HCl   ------------> KCl + H2O

millimoles of KOH = millimoles of HCl

0.240 x 27.96 = 96.34 x M

M = 0.0696

Molarity of HCl = 0.06964 M

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