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20 June, 22:28

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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  1. 21 June, 00:49
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    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
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