Answer to Question #147012 in General Chemistry for Marie

Question

A sample of potassium phthalate, KHC8H4O4, monobasic, weighing 4.0700g is titrated with NaOH solution and back titrated with HCl. NaOH required = 46.40 mL; HCl required = 5.35mL. If each mL of the HCl is equivalent to 0.01600g of Na2O, what volume of H2O or of 6.00N NaOH must be added to 500mL of the above NaOH to bring it to 0.5000N?

Answer 1

2HCl_(aq)+Na₂O_(s)=H₂O_(l)+2NaCl_(aq)

Moles of Na₂O= 0.016/62=.000258moles

Moles of HCl= 0.000258×2

                     =0.000516moles

Total moles of acid in 5.35 ml= 5.35×0.000516

                                                =0.0027606moles

HCl_(aq)+NaOH_(aq)=H₂O_(l)+NaCl_(aq)

Moles of NaOH= 0.0027606moles

Molarity=1000×0.00027606/46.4

            =0.0595M

C₁V₁=C₂V₂

46.40×0.0595=500C₂

C₂= 0.0055216M

Molarity= Normality/Equivalent

For 0.5000N NaOH=0.5/1

                               =0.5M

For 6.000M NaOH= 6/1 =6M

C₁V₁=C₂V₂

C₁= (6+0.0055216) =6.0055216M

V₁= 500mL

C₂= 0.5M

V₂=?

6.0055216×500=0.5 V₂

V₂= 6005.5216mL

   =6.006L

Volume of water added= (6.006-0.5) = 5.506L