If it requires milliliters of .045 molar barium hydroxide to neutralize 38.5 milliliters of nitric acid, solve for the molarity of the nitric acid solution. unbalanced equation: ba(oh)2+hno3-> ba(no3)2+h2o

1) It is missing the amount (mililiters) of the 0.045 molar barium hydroxide.

2) I will show you how to solve this using a generic value V, for this volumen. And, then I will calculate the requested molarity for some assumed values of V.

3) Start with stating the balanced chemical equation:

Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O

4) State the mole ratio between the two reactants:

1 mol Ba(OH)₂ : 2 mol HNO₃

5) Set the equations for the number of moles of both solutions

M = n / V ⇒ n = M × V

⇒ number of moles of baryum hidroxide: 0.045V

number of moles of nitric acid: 0.0385 M

6) As per the theoretical mole ratio, the neutralization implies number of moles of barium hydroxide equals 2 times the number of moles of nitric acid:

⇒ 0.045V = 2×0.0385×M

And, solve for M: M = 0.045V / 0.077 = 0.5844V

That is the fomula that you can use: M = 0.5844V.

7) Here some examples, for different values of the volume V.

Volum in ml      V (in liters)       M = 0.5844V

10                      10/1000              0.0058

20                       20/1000              0.012

30                        30/1000             0.018

38.5                      38.5 / 1000        0.023

77                         77/1000             0.045

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