Question

A solution is made by adding 0.380 g Ca(OH)2(s), 50.0 mL of 1.20 M HNO3, and enough water to make a final volume of 75.0 mL.

Part A

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

Express your answer using three decimal places.

pH =

Answer 1

Let's determine the moles of each reactant:

moles Base = 0.380 g / (40+32+2) = 0.00514 moles

moles Acid = 1.20 mol/L * 0.050 L = 0.06 moles

The reaction would be:

2HNO₃ + Ca(OH)₂ ----------> Ca(NO₃)₂ + 2H₂O

0.06 0.00514

From here the limitant reactant is the base, so the moles remaining of acid would be:

moles of acid remaining = 0.06 - 0.00514 = 0.05486 moles

Now, let's calculate the concentration:

M = 0.05486 / 0.075 = 0.7315 M

HNO₃ is a strong acid, and because of that, it dissociates completely in solution so:

pH = -log(0.7315)

pH = 0.14

EDIT:

The procedure above is the correct one, but you may use the concentration of the solution:

[M] = 0.00514 / 0.075 = 0.0685 M

pH = -log(0.0685) = 1.16

Hope this helps