Use the Ref to access im values if needed for this q A student is asked to stand

Question

Use the Ref to access im values if needed for this q A student is asked to standardize a solution of sodium hydroxide. He weighs out 1,06 g potassium hydrogen phthalate (KHCgH404, treat this as a monoprotic acid). It requires 28.4 mL of sodium hydroxide to reach the endpoint. A. What is the molarity of the sodium hydroxide solution?M This sodium hydroxide solution is then used to titrate an unknown solution of hydroiodic acid. B. If 13.8 mL of the sodium hydroxide solution is required to neutralize 26.7 mL of hydroiodic acid, what is the molarity of the hydroiodic acid solution?M Submit Answer Retry Entire Group 4 more group attempts remaining

Explanation / Answer

KHC8H4O4 + NaOH ------------> NaKC8H4O4 + H2O

no of moles of KHP = W/G.M.Wt

                                 = 1.06/204.22   = 0.00519 moles

1 mole of KHC8H4O4 react with 1 mole of NaOH

0.00519 moles of KHC8H4O4 react with 0.00519 moles of NaOH

volume of NaOH = 28.4ml   = 0.0284L

no of moles of NaOH = molarity * volume in L

                 0.00519    = molarity* 0.0284

molarity                     = 0.00519/0.0284   = 0.183 M

The molarity of NaOH = 0.183 M
NaOH + HCl ----------------------> NaCl + H2O

1 mole   1mole

HCl                                                                      NaOH

M1 =                                                                M2 = 0.183M

V1 = 26.7ml                                                     V2   = 13.8ml

n1 = 1                                                              n2 =1

                 M1V1/n1    =       M2V2/n2

                     M1         =       M2V2n1/V1n2

                                    =     0.183*13.8*1/26.7*1

                                      = 0.0946M

The molarity of HCl       = 0.0946M

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