The contents of the flask are transferred quantitatively to a 2L volumetric. 100ml of 1 M HCL is added and the flask filled to the mark with water. There is still observable solid calcium hydroxide in the bottom of the new flask. What is the pH of this solution?
A) 2.94
B) 12.14
C) 2.16
D) 7.00
E) 11.83

Question

Quizplus · Accepted Answer

Calcium hydroxide (Ca(OH)2) is a strong base and reacts with hydrochloric acid (HCl) to form calcium chloride (CaCl2) and water (H2O) according to the following balanced equation:

Ca(OH)2 + 2HCl → CaCl2 + 2H2O

Since the contents of the flask were transferred quantitatively to a 2L volumetric flask, the number of moles of calcium hydroxide in the flask is the same as the number of moles of calcium hydroxide in the 2L volumetric flask. Therefore, we can use the following equation to calculate the initial number of moles of calcium hydroxide:

moles of Ca(OH)2 = Molarity of Ca(OH)2 × Volume of Ca(OH)2

= 0.05 M × 0.025 L

= 0.00125 moles

Next, we can use the balanced equation to calculate the number of moles of hydrochloric acid that reacted with the calcium hydroxide:

1 mole of Ca(OH)2 reacts with 2 moles of HCl

Therefore, the number of moles of HCl that reacted = (0.00125 moles of Ca(OH)2) × (2 moles of HCl/1 mole of Ca(OH)2) = 0.0025 moles of HCl

The total volume of the solution in the 2L volumetric flask is 2L, which is equivalent to 2000 ml. So, we can use the following equation to calculate the final concentration of the solution:

M1V1 = M2V2

Where M1 is the initial concentration of the HCl solution (1M), V1 is the volume of the HCl solution added (100ml or 0.1L), M2 is the final concentration of the solution, and V2 is the final volume of the solution (2L or 2000ml).

Substituting the values:

(1 M) × (0.1 L) = (M2) × (2 L)

M2 = 0.05 M

Now we can use the equilibrium constant expression for the dissociation of water to calculate the pH of the solution:

Kw = [H3O+][OH-] = 1.0 × 10^-14

At 25°C, the concentration of H3O+ and OH- in pure water is equal to 1.0 × 10^-7 M. Since the solution is basic due to the presence of excess hydroxide ions from the reaction of calcium hydroxide with hydrochloric acid, we can assume that the concentration of OH- ions in the solution is much greater than 1.0 × 10^-7 M. Let x be the concentration of H3O+ ions in the solution. Then,

Kw = [H3O+][OH-] = x[OH-] = 1.0 × 10^-14

[OH-] = 1.0 × 10^-14 / x

Assuming [OH-] >> [H3O+], we can simplify this expression to:

[OH-] ≈ 1.0 × 10^-14

Now we can use the equation for the concentration of hydroxide ions in the solution:

moles of OH- = Molarity of OH- × Volume of solution

moles of OH- = (0.05 M) × (2 L)

moles of OH- = 0.1 moles

Converting moles of OH- to concentration:

[OH-] = moles of OH- / Volume of solution

[OH-] = 0.1 moles / 2 L

[OH-] = 0.05 M

Substituting this value into the equation for the dissociation of water:

Kw = [H3O+][OH-] = x(0.05 M) = 1.0 × 10^-14

x = 2.0 × 10^-13

Therefore, the concentration of H3O+ ions in the solution is 2.0 × 10^-13 M. Taking the negative logarithm of this concentration gives:

pH = -log[H3O+]

pH = -log(2.0 × 10^-13)

pH = 12.14

Therefore, the pH of the solution is 12.14, and the correct answer is B.

The Contents of the Flask Are Transferred Quantitatively to a 2L