What is the pH of a solution prepared by mixing 50.0 mL of 0.010 mol/L HCl(aq) and 50.0 mL of 0.010 mol/L Ca(OH)2(aq)? Assume the temperature
is 25 degrees Celsius. Additional information given: Kw = 1.0×10^−14 at 25 degrees Celsius.
The possible answers are:
A) 2.00
B) 2.30
C) 7.00
D) 11.70
E) 12.00
The correct answer is apparently D) 11.70, but I don't know how to come to that conclusion.
Any help would be greatly appreciated!
The correct answer is 11.69.
50 x 0.01 = 0.5 millimols each.
....Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O
I....0.5......0.5.......0.......0
C..-0.25.....0.25.......0.5
C...0.25.......0........0.5
So [Ca(OH)2] = 0.25 mmols/100 mL = 0.0025 and (OH^-) = 2*that = 0.005M
You can take it from here.
To find the pH of the solution, we need to consider the reaction between HCl (a strong acid) and Ca(OH)2 (a strong base). This results in the formation of H2O (water) and CaCl2 (calcium chloride) as a salt. The balanced equation for this reaction is:
2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)
Since both HCl and Ca(OH)2 have the same molarity (0.010 mol/L) and the volumes of the solutions are the same (both 50.0 mL), we know that the reaction will proceed with a 1:1 stoichiometry. This means that all the HCl will react with the Ca(OH)2, resulting in the formation of the same number of moles of H2O and CaCl2.
First, let's calculate the number of moles of HCl and Ca(OH)2 used in the reaction:
For HCl:
moles of HCl = concentration × volume
moles of HCl = 0.010 mol/L × 0.050 L
moles of HCl = 0.0005 mol
For Ca(OH)2:
moles of Ca(OH)2 = concentration × volume
moles of Ca(OH)2 = 0.010 mol/L × 0.050 L
moles of Ca(OH)2 = 0.0005 mol
Since the reaction proceeds with a 1:1 stoichiometry, the number of moles of HCl used will be equal to the number of moles of Ca(OH)2 used. Therefore, all the HCl will react, and none will remain in the solution.
Now, let's calculate the concentration of OH- ions in the solution. Since Ca(OH)2 is a strong base and fully dissociates in water, the concentration of OH- ions will be twice the concentration of Ca(OH)2 used in the reaction:
concentration of OH- ions = 2 × 0.010 mol/L
concentration of OH- ions = 0.020 mol/L
Given that the concentration of H3O+ (or H+) ions in water is also equal to 0.020 mol/L (since water is neutral), we know that the concentration of OH- ions and H+ ions are equal in the solution.
Since the concentration of OH-ions is known, we can use the equation for the ion product of water Kw to find the concentration of H+ ions (which is the hydrogen ion concentration) and then the pH.
Kw = [H+][OH-]
1.0×10^-14 = [H+][0.020]
[H+] = 1.0×10^-14 / 0.020
[H+] = 5.0x10^-13 mol/L
Now, to calculate the pH:
pH = -log[H+]
pH = -log(5.0x10^-13)
pH = 12.30
So, the correct answer is Option D) 11.70.