In a reaction 25.00mL of nitric acid required 51.4mL of 0.73M calcium hydroxide. Calculate the grams of nitric acid per liter of solution

47.27 g HNO₃

Explanation:

The reaction between nitric acid (HNO₃) and calcium hydroxide (Ca(OH)₂) is the following:

2HNO₃(aq) + Ca(OH)₂(aq) → Ca(NO₃)₂ + 2H₂O(l)

According to the chemical equation, 2 moles of HNO₃ reacts with 1 mol of Ca(OH)₂. At the point of total neutralization, the total number of moles of  HNO₃ reacts with the total number of moles of Ca(OH)₂:

2 x moles HNO₃ = moles Ca(OH)₂

The number of moles is equal to the product of molarity of the solution (M) and the volume in liters (V):

M(HNO₃) = ?

M(Ca(OH)₂) = 0.73 M

V(HNO₃) = 25.00 mL x 1 L/1000 mL = 0.025 L

V(Ca(OH)₂) = 51.4 mL x 1 L/1000 mL = 0.0514 L

So, we can write the equation at neutralization point as:

2 x (M(HNO₃) x V(HNO₃)) = M(Ca(OH)₂) x V(Ca(OH)₂)

From this, we can calculate the molarity of HNO₃:

M(HNO₃) = (M(Ca(OH)₂) x V(Ca(OH)₂))/2 x V(HNO₃)

               = (0.73 M x 0.0514 L)/(2 x 0.025 L)

               = 0.75 M HNO₃

We can convert the molarity from M (mol/L) to g/L by using the molecular weight of HNO₃:

Mw(HNO₃) = 1 g/mol H + 14 g/mol N + (3 x 16 g/mol O) = 63 g/mol

Finally, we multiply the molarity by the molecular weight to obtain the grams of nitric acid per liter of solution:

0.75 mol/L HNO₃ x 63 g/mol = 47.27 g HNO₃

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