What is the pH of the resulting solution?

Question

For his titration experiment, Alex determines that the equivalence point is reached when an unknown concentration of a 75.0 mL sample of nitric acid is added with 92.5 mL of 1.0 mol/L calcium hydroxide solution. If only 50.0 mL of 1.0 mol/L Ca(OH)2 were added to the nitric acid solution, what would the pH of the resulting solution be?

Answer 1

Let us first look at the balanced equation:

2HNO₃ + Ca(OH)₂ ==> Ca(NO₃)₂ + 2H₂O

From the information given, we can find the concentration of the HNO₃ as follows:

moles of Ca(OH)₂ used for equivalence = 0.0925 L x 1.0 mol/L = 0.0925 mol Ca(OH)₂

moles of HNO₃ present in 75,0 ml = 0.0925 mol Ca(OH)₂ x 2 mol HNO₃/mol Ca(OH)₂ = 0.185 mol HNO₃

Concentration (M) of HNO₃ = 0.185 mol/0.075 L = 2.47 mol/L = 2.47 M HNO₃

Next, we use this information to find the pH after addition of Ca(OH)₂. Recall that pH is -log [H⁺], so we first need to find [H⁺]. Again, from the balance equation 2HNO₃ + Ca(OH)₂ ==> Ca(NO₃)₂ + 2H₂O...

moles Ca(OH)₂ = 0.050 L x 1.0 mol/L = 0.050 mol Ca(OH)₂ present

moles HNO₃ present = 0.075 L x 2.47 mol/L = 0.185 mol HNO₃

moles HNO₃ used up = 0.050 mol Ca(OH)₂ x 2 mol HNO₃/mol Ca(OH)₂ = 0.10 mol HNO₃ used up

moles HNO₃ remaining = 0.185 mol - 0.1 = 0.085 mol HNO₃ left over

Total volume = 75 ml + 50 ml = 125 ml = 0.125 L

Final [HNO₃] = [H⁺] = 0.085 mol/0.125 L = 0.68 M HNO₃

pH = -log 0.68

pH = 0.17