A 1.00 g sample of NaOH which has been exposed to the air  for  some time is dissolve in water and water and diluted to 500.0 ml.   100.0 ml of the solution is titrated with 0.1062 N HCl  (MOH indicator) and requires 38.60 ml. A second 100 .0 ml portion  is titrated with excess BaCl2,  diluted to 250 .0 ml and filtered.   200.0 ml of this filtrate then require  29.62 ml of the HCl  ( HpH  indicator) .  Calculate the %  NaOH  and Na2CO3  in the original sample.

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Q: A 50.0 mL solution of Ca(OH)₂ with an unknown concentration was titrated with 0.340 M HNO₃. To reach…

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A: Given:  Mass of sample = 188.5 mg  Molarity of HCl solution = 0.1065 M  Volume of HCl solution…

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