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A sample of potassium hydrogen oxalate, KHC204, weighing 0.717 g, was dissolved in water and titrated with 18.47 mL of an NaOH...Asked by mary
A sample of potassium hydrogen oxalate, KHC2O4, weighing 0.717 g, was dissolved in water and titrated with 18.47 mL of an NaOH solution. Calculate the molarity of the NaOH solution.
Answers
Answered by
Devron
First, determine the molarity of KHC2O4:
0.717g*(1 mol/128.12g)=moles of KHC2O4
18.47 mL=0.01847L
Molarity of KHC2O4=moles of KHC2O4/0.01847L
The reaction is a 1 mole to 1 mole reaction. Therefore, moles of KHC2O4=moles of NaOH.
So, Molarity of KHC2O4=Molarity of NaOH
***Answer contains three significant figures.
0.717g*(1 mol/128.12g)=moles of KHC2O4
18.47 mL=0.01847L
Molarity of KHC2O4=moles of KHC2O4/0.01847L
The reaction is a 1 mole to 1 mole reaction. Therefore, moles of KHC2O4=moles of NaOH.
So, Molarity of KHC2O4=Molarity of NaOH
***Answer contains three significant figures.
Answered by
Marion
0.185
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