. A student was conducting a titration experiment and was asked to calculate the

Question

. A student was conducting a titration experiment and was asked to calculate the theoretical pH of the solution at various points in the titration curve. In this experiment, the student added 40.75 ml of a 0.39 M solution of HCl to a 200 ml beaker. The student then added 39.7 ml of deionized water. After adding the water to the solution, the mixture was stirred to ensure to ensure proper mixing of solution. Next, the solution in the beaker was titrated with 22.37 ml of 0.39 M NaOH. Note that this is before the equivalence point. a) How many moles of HCI was in the intial solution of 40.75 ml of 0.39 M HCI? b) What is the molarity of the inital HCl solution of 40.75 ml of 0.39 M HCI afer adding 39.7 ml the deionized water? c) How many moles of NaOH was titrated (added) into the beaker? d) How many moles of unreacted HCl is left in the beaker after addition of 22.37 ml of 0.39 M NaOH is added. e) What is the total volume of the solution in liters at the end of this titration? f What would be the theoretical pHH at this point?

Explanation / Answer

a) Moles of HCl = Molarity× Volume of solution(in L)= (0.39×40.75)/1000=0.0158 moles

b) After adding deionised water, volume of solution increases=40.75+39.7=80.45ml

Now, moles of HCl=molarity×volume of solution(in L)=(0.39×80.45)/1000=0.0313 moles

C) moles of NaOH added into beaker= molarity of NaOH×volume of solution(in L)=(0.39×22.37)/1000=0.00872 moles

D) neutratization reaction takes place between HCl and NaOH

Moles of HCl=0.0313

Moles of NaOH=0.00772

Moles of HCl left unreacted=0.0313-0.00772=0.02358 moles

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