Assuming complete dissociation, what is the pH of a 3.73 mg/L Ba(OH)2 solution?
To find the pH of a Ba(OH)2 solution, first, let's calculate the concentration of hydroxide ions (OH-) in the solution.
The molar mass of Ba(OH)2 is:
Ba = 137.33 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
So, the molar mass of Ba(OH)2 is:
137.33 + (16.00 * 2) + (1.01 * 2) = 171.35 g/mol
Now, we need to calculate the number of moles of Ba(OH)2 in 3.73 mg/L:
3.73 mg/L = 3.73 * 10^-6 g/mL
To convert grams to moles, divide the mass by the molar mass:
(3.73 * 10^-6 g/mL) / 171.35 g/mol = 2.17 * 10^-8 mol/mL
Since the solution is assumed to be completely dissociated, the concentration of hydroxide ions (OH-) is twice the concentration of Ba(OH)2:
2 * (2.17 * 10^-8 mol/mL) = 4.34 * 10^-8 mol/mL
Now, we can calculate the pOH of the solution:
pOH = -log10[OH-]
pOH = -log10[4.34 * 10^-8]
Using a calculator, the pOH is approximately 7.363.
Finally, to find the pH, we can use the equation:
pH = 14 - pOH
pH = 14 - 7.363
pH is approximately 6.637.
Therefore, the pH of the 3.73 mg/L Ba(OH)2 solution is approximately 6.637.
To determine the pH of a Ba(OH)2 solution, we need to consider the dissociation of the compound. Ba(OH)2 dissociates into Ba2+ ions and OH- ions in an aqueous solution.
Given that the compound is assumed to completely dissociate, we can assume that the concentration of Ba2+ ions is twice the concentration of Ba(OH)2 and the concentration of OH- ions is twice the concentration of Ba(OH)2. This is because for every one Ba(OH)2 molecule, two Ba2+ ions and two OH- ions are produced.
First, we need to calculate the concentration of Ba(OH)2 in moles per liter (mol/L):
1. Convert the given mass concentration of Ba(OH)2 from mg/L to g/L:
- 3.73 mg/L × (1 g/1000 mg) = 0.00373 g/L
2. Calculate the molar mass of Ba(OH)2:
- Ba: Atomic mass = 137.33 g/mol
- O: Atomic mass = 16.00 g/mol
- H: Atomic mass = 1.01 g/mol
- Molar mass of Ba(OH)2 = (137.33 g/mol) + 2 * [(16.00 g/mol) + (1.01 g/mol)] = 137.33 g/mol + 34.02 g/mol = 171.35 g/mol
3. Convert the mass concentration of Ba(OH)2 from g/L to mol/L:
- 0.00373 g/L × (1 mol/171.35 g) = 2.18 × 10^(-5) mol/L
Since Ba(OH)2 completely dissociates, the concentration of Ba2+ ions and OH- ions is twice the concentration of Ba(OH)2:
Concentration of Ba2+ ions = 2 × 2.18 × 10^(-5) mol/L = 4.36 × 10^(-5) mol/L
Concentration of OH- ions = 2 × 2.18 × 10^(-5) mol/L = 4.36 × 10^(-5) mol/L
Now, we can calculate the pOH of the solution:
pOH = -log10[OH-] = -log10(4.36 × 10^(-5)) = 4.36
Since pH + pOH = 14, we can calculate the pH of the solution:
pH = 14 - pOH = 14 - 4.36 = 9.64
Therefore, the pH of a 3.73 mg/L Ba(OH)2 solution, assuming complete dissociation, is approximately 9.64.