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I have two question i cant get and i need help with them bytonight please! 1. 24

ID: 684495 • Letter: I

Question

I have two question i cant get and i need help with them bytonight please! 1. 24.60 ml of 0.185 HNO3 are titrated with 27.35 ml of a KOHsolution. What is the molarity of the KOH? 2. The results of a titration show that 20.34 ml of 0.1003Moxalic acid (H2C2O4) solution exactly neutralize 25.00 ml of asodium hydroxide solution. What is the molarity of the NaOHsolution? (Hint: write the chemical equation for the reactionfirst. Both hydrogens in H2C2O4 are neutralized.) I have two question i cant get and i need help with them bytonight please! 1. 24.60 ml of 0.185 HNO3 are titrated with 27.35 ml of a KOHsolution. What is the molarity of the KOH? 2. The results of a titration show that 20.34 ml of 0.1003Moxalic acid (H2C2O4) solution exactly neutralize 25.00 ml of asodium hydroxide solution. What is the molarity of the NaOHsolution? (Hint: write the chemical equation for the reactionfirst. Both hydrogens in H2C2O4 are neutralized.)

Explanation / Answer

1. MaVa =MbVb        0.185M*24.60mL = Mb*27.35mL          Mb =(0.185M*24.60mL)/27.35mL                = 0.166M Mb , molarity of base is 0.166M 2.H2C2O4(aq)   +NaOH (aq) ..............>Na2C2O4 (aq) + 2H2O(l) MaVa/na = MbVb/ nb    (0.1003M*20.34mL)/1 = (25.00mL*Mb)/2      ((2.040102M/mL)*2)/25.0mL = Mb          Mb = 0.163M Molarity of reqiured NaOH solution is 0.163M Molarity of reqiured NaOH solution is 0.163M