Question

After standardizing your NaOH solution, you have determined that the concentration is actually 0.1125 M. What is the weight percent concentration of KHP (MW=204.23) in your unknown sample of mass 1.625 g if it takes 38.40 mL of titrant to reach the equivalence point?

A student weighs out 1.7500 grams of dried potassium hydrogen phthalate standard (MW=204.23) and finds that it takes 39.05 mL to reach the endpoint of the titration with the solution of NaOH. What is the molarity of the NaOH titrant?

After standardizing your NaOH solution, you have determined that the concentration is actually 0.1125 M. If your unknown contains a 35.60 weight percent concentration of KHP (MW=204.23) in your unknown sample of mass 1.500 g how many milliliters of titrant will be needed to reach the equivalence point of the titration?

Answer #1

**KHP + NaOH ----> salt + H2O**

**No of mol of NaOH = 38.4/1000*0.1125 = 0.00432
mol**

**No of mol of KHP reacted = 0.00432 mol**

**mass of kHP reacted = 0.00432*204.23 = 0.88
grams**

**mass percentage = 0.88 / 1.625*100 =
54.15%**

**No of mol of KHP reacted = 1.75/204.23 = 0.00857
mol**

**Molarity of NaOH = 0.00857/39.05*1000 = 0.22
M**

**No of mol of KHP = 1.5*(35.6/100)*(1/204.23) = 0.0026
mol**

**No of mol of NaOh = 0.0026 mol**

**volume of NaOH needed = 0.0026 / 0.1125 = 23.11
ml**

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