Question

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?

Answer 1

The pH of the final solution, after 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, is approximately 13.779.

Step-by-step explanation:

The pH of the final solution can be calculated by considering the reaction between Ca(OH)2 and HNO3. First, calculate the moles of HNO3 by multiplying the volume (45.0 mL) by the molarity (1.00 M), which gives you 0.0450 moles of HNO3.

Next, use the balanced chemical equation to determine the moles of Ca(OH)2, which is half the moles of HNO3: 0.0225 moles. Since the solution volume is 75.0 mL, or 0.0750 liters, you can calculate the concentration of Ca(OH)2 to be 0.0225 moles / 0.0750 L = 0.300 M.

To find the pH, you can use the equation pH = -log[H+]. Since Ca(OH)2 is a strong base and completely dissociates into two OH- ions, the concentration of OH- ions is twice the concentration of Ca(OH)2: 0.600 M. Taking the negative logarithm of 0.600 M gives you a pOH of 0.221.

Subtracting the pOH from 14 gives you the pH of the solution, which is approximately 13.779.

Answer 2

0.323

Step-by-step 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