Plz . i will give if it requires 18.2 milliliters of 0.45 molar barium hydroxide to neutralize 38.5 milliliters of nitric acid, solve for the molarity of the nitric acid solution. show all of the work used to solve this problem. unbalanced equation: ba(oh)2 + hno3 ba(no3)2 + h2o

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!