Respuesta :
Answer: The pH of the solution is 9.68
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
We are given:
Mass of solute (barium hydroxide) = 4.02 mg = 0.00402 g (Conversion factor: 1 g = 1000 mg)
Molar mass of barium hydroxide = 171.34 g/mol
Volume of solution = 1 L
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{0.00402g}{171.34g/mol\times 1L}\\\\\text{Molarity of solution}=2.4\times 10^{-5}M[/tex]
To calculate the pH of the solution, we need to determine pOH of the solution. To calculate pOH of the solution, we use the equation:
[tex]pOH=-\log[OH^-][/tex]
On complete dissociation, 1 mole of barium hydroxide produces 2 moles of hydroxide ions
We are given:
[tex][OH^-]=4.8\times 10^{-5}M[/tex]
Putting values in above equation, we get:
[tex]pOH=-\log(4.8\times 10^{-5})\\\\pOH=4.32[/tex]
To calculate pH of the solution, we use the equation:
[tex]pH+pOH=14\\pH=14-4.32=9.68[/tex]
Hence, the pH of the solution is 9.68
Considering the definition of pH, pOH and strong base, the pH of a 4.02 mg/L Ba(OH)₂ solution is 9.67.
You have a 4.02 mg/L Ba(OH)₂ solution. Being the molar mass (that is, the mass of one mole of a substance, which can be an element or a compound.) of barium hydroxide 171.34 g/mole, and being 1 mg=0.001 g then, the amount of moles that contain 4.02 mg of Ba(OH)₂ can be calculated as:
[tex]4.02 mgx\frac{0.001 grams}{1 mg} x\frac{1 mole}{171.34 grams} =[/tex] 2.35×10⁻⁵ moles
Then, the concentration of the Ba(OH)₂ solution is 2.35×10⁻⁵[tex]\frac{moles}{L}[/tex]=2.35×10⁻⁵ M.
On the other side, a Strong Base is that base that in an aqueous solution completely dissociates into the cation and hydroxide ion.
In this case, Ba(OH)₂ is a strong base. Then, the dissociation reaction will be:
Ba(OH)₂ → Ba²⁺ + 2 OH⁻
Since Ba(OH)₂ will completely dissociate in water, you can observe that the concentration of OH⁻ will be twice the concentration of Ba(OH)₂.
So, [OH⁻]= 2×2.35×10⁻⁵ M
[OH⁻]= 4.7×10⁻⁵ M
pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:
pOH= - log [OH⁻]
So, in this case:
pOH= - log (4.7×10⁻⁵ M)
pOH= 4.33
pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.
The following relationship can be established between pH and pOH:
pH + pOH= 14
Being pOH= 4.33, pH is calculated as:
pH + 4.33= 14
pH= 14 - 4.33
pH= 9.67
Finally, the pH of a 4.02 mg/L Ba(OH)₂ solution is 9.67.
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