Question

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What is the molarity of the nitric acid if 35.26 mL of 0.2360 M Mg(OH)2 solution is required to completely neutralize 36.08 mL of the nitric acid solution?

(Hint: balance the chemical equation first.)

Answer 1

The balanced chemical equation for the reaction between nitric acid (HNO3) and magnesium hydroxide (Mg(OH)2) is:

2HNO3 + Mg(OH)2 → Mg(NO3)2 + 2H2O

From the equation, we can see that 2 moles of nitric acid react with 1 mole of magnesium hydroxide. Therefore, the number of moles of nitric acid is:

n(HNO3) = (0.2360 M) x (0.03526 L) = 0.008325 mol

According to the stoichiometry of the balanced equation, this amount of nitric acid would react with half as many moles of magnesium hydroxide, or:

n(Mg(OH)2) = 0.004163 mol

The molarity of the magnesium hydroxide solution is:

M(Mg(OH)2) = n/V = 0.004163 mol / 0.03608 L = 0.1153 M

Since the number of hydroxide ions produced by the reaction is twice the number of moles of magnesium hydroxide, the hydroxide ion concentration is:

[OH-] = 2 x 0.004163 mol / 0.03608 L = 0.2311 M

Finally, the pH of the solution can be calculated using the following equation:

pH = 14 - log([OH-]) = 14 - log(0.2311) ≈ 12.64

Answer 2

The balanced chemical equation for the reaction between nitric acid (HNO3) and magnesium hydroxide (Mg(OH)2) is:

2HNO3 + Mg(OH)2 → Mg(NO3)2 + 2H2O

From the equation, we can see that 2 moles of nitric acid react with 1 mole of magnesium hydroxide. Therefore, the moles of nitric acid in the solution can be calculated as:

moles of HNO3 = (0.2360 mol/L) x (0.03526 L) = 0.008329 mol

Since 2 moles of HNO3 react with 1 mole of Mg(OH)2, we can calculate the moles of Mg(OH)2 by:

moles of Mg(OH)2 = 0.008329 mol / 2 = 0.004165 mol

Now we can calculate the molarity of the nitric acid solution by dividing the moles of HNO3 by the volume of the solution in liters:

Molarity of HNO3 = moles of HNO3 / volume of HNO3 solution in liters

The volume of the nitric acid solution used is 36.08 mL, which is equivalent to 0.03608 L. Therefore:

Molarity of HNO3 = 0.008329 mol / 0.03608 L = 0.2305 M

Therefore, the molarity of the nitric acid solution is 0.2305 M.