If it requires 38.5 milliliters of 0.85 molar nitric acid to neutralize 20.0 milliliters of barium hydroxide, solve for the molarity of barium hydroxide. show all of the work used to solve this problem.. unbalanced equation: ba(oh)2 + hno3 ba(no3)2 + h2o. ( i already balanced the equation)

Molarity HNO3 = 0.85 M

Volume HNO3 = 38.5 mL in liters: 38.5 / 1000 => 0.0385 L

Molarity Ba(OH)2 = ?

Volume Ba(OH)2 = 20.0 mL / 1000 => 0.02 L

Number of moles HNO3 :

n = M x V

n = 0.85 x 0.0385 => 0.032725 moles

mole ratio :

Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O

1 mole Ba(OH)2 2 moles HNO3
( moles Ba(OH)2 )  0.032725 HNO3

moles Ba(OH)2 = 0.032725 x 1 / 2

moles Ba(OH)2 = 0.032725 / 2

= 0.0163625 moles of Ba(OH)2

Molarity Ba(OH)2 :

M = n / V

M =0.0163625 / 0.02

= 0.82 M

hope this helps!

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