A solution is made by adding 0.350 g Ca (OH) 2 (s), 45.0 mL of 1.00 M HNO3, and enough water to make a final volume of 75.0 mL. Assuming that all of the solid dissolves, what is the pH of the final solution?

Question

Jaylin Parsons

Chemistry

3 November, 16:42

A solution is made by adding 0.350 g Ca (OH) 2 (s), 45.0 mL of 1.00 M HNO3, and enough water to make a final volume of 75.0 mL. Assuming that all of the solid dissolves, what is the pH of the final solution?

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Answers (1)

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Matthias Wright · Answer 1

0.323

Explanation:

number of moles of Ca (OH) 2 (s) = mass given / molar mass = 0.350 g / 74.093 g/mol = 0.00472 mol

number of mole of HNO3 = Molarity * volume in Liters = 1.0 * (45 / 1000) = 0.045 M

Ca (OH) 2 (s) + 2 HNO3 (aq) → Ca (NO₃) ₂ (aq) + 2 H₂O (I)

1 mole of calcium hydroxide react with 2 mole of trioxonitrate (V)

0.00472 mole will require 0.00945 mole

but we have 0.045 mole of the acid

net mole = 0.045 - 0.00945 = 0.0356 mole

Molarity of the net mole = 0.0356 / (75/1000) = 0.475 M

pH = - log (0.475) = 0.323