Question

A solution is made by adding 0.330 g Ca(OH)2(s), 40.0 mL of 1.00 M HNO3, and enough water to make a final volume of 75.0 mL.

A)Assuming that all of the solid dissolves, what is the pH of the final solution?

Answer 1

Given,

Mass of Ca(OH)₂(s) = 0.330 g

Concentration of HNO₃ = 1.00 M

Volume of HNO₃ = 40.0 mL x ( 1L/1000 mL) = 0.04 L

Now, the reaction between Ca(OH)₂ and HNO₃ is,

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

Calculating the number of moles of Ca(OH)₂ and HNO₃ from the given data,

= 0.330 g Ca(OH)₂ x (1 mol/ 74.093 g)

= 0.004454 mol Ca(OH)₂

Now,

= 1.00 M x 0.04 L

= 0.04 mol HNO₃

Now, calculating the number of moles of HNO₃ required to react completely with the given moles of Ca(OH)₂,

= 0.004454 mol Ca(OH)₂ x ( 2 mol HNO₃ / 1 mol Ca(OH)₂)

= 0.0089 mol HNO₃ required to react completely with the given moles of Ca(OH)₂

Similarly, calculating the number of moles of Ca(OH)₂ required to react completely with the given moles of HNO₃,

= 0.04 mol HNO₃ x ( 1 mol Ca(OH)₂ / 2 mol HNO₃)

= 0.02 mol Ca(OH)₂ required to react completely with the given moles of HNO₃

The moles of Ca(OH)₂ required to react completely with the given moles of HNO₃ are more than the given moles of Ca(OH)₂, thus, Ca(OH)₂ is the limiting reactant.

Now, calculating the excess moles of HNO₃,

= Initial moles of HNO₃ - Moles of HNO₃ used to react with Ca(OH)₂,

= 0.04 mol -0.0089 mol

= 0.03109 mol HNO₃

Now, we know the dissociation of HNO₃ in aqueous solution,

HNO₃(aq) + H₂O(l) NO₃^-(aq) + H₃O⁺(aq)

From the excess moles of HNO₃ calculated and the mole ratio, calculating the moles of H₃O⁺,

= 0.03109 mol HNO₃ ​​​​​​​ x ( 1 mol H₃O⁺ / 1 mol HNO₃)

= 0.03109 mol H₃O⁺

The final volume of the solution = 75.0 mL x ( 1 L /1000 mL) = 0.075 L

Thus,

[H₃O⁺] = 0.03109 mol / 0.075

[H₃O⁺] = 0.4146 M

We know, the formula to calculate pH,

pH = -log[H₃O⁺]

pH = -log[0.4146]

pH = 0.382