Respuesta :
Given:
Concentration of Ba(OH)2 = 0.348 M
To determine:
pOH of the above solution
Explanation:
Based on the stoichiometry-
1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion
Therefore, concentration of OH- ion = 2*0.348 = 0.696 M
pOH = -log[OH-] = - log[0.696] = 0.157
Ans: pOH of 0.348M Ba(OH)2 is 0.157
0.157
Further explanation
Given:
0.348 M of Ba(OH)₂ solution at 25⁰C
Question:
Determine the pOH of Ba(OH)₂.
The Process:
Ba(OH)₂ represents a strong base. The characteristic of a strong base is it has no ionization constant, Kb or the degree of ionization, α. Ba(OH)₂ is fully ionized in water.
The concentration of OH⁻ ions will determine the pOH value and pH by looking at the number of OH⁻ ions or sometimes referred to as base valence.
Let us prepare the concentration of OH⁻ ions.
[tex]\boxed{ \ Ba(OH)_2_{(aq)} \rightarrow Ba^{2+}_{(aq)} + 2OH^-_{(aq)} \ }[/tex]
[tex]\boxed{ \ [OH^-] = \frac{2}{1} \times [Ba(OH)_2] \ }[/tex]
[tex]\boxed{ \ [OH^-] = 2 \times 0.348 \ M \ }[/tex]
∴ [tex]\boxed{ \ [OH^-] = 0.696 \ M \ }[/tex]
After obtaining the concentration of OH⁻ ions, we can immediately determine the pOH value.
[tex]\boxed{ \ pOH = -log[OH^-] \ }[/tex]
[tex]\boxed{ \ pOH = -log \ 0.696 \ }[/tex]
Thus, we get a pOH of 0.157.
_ _ _ _ _ _ _ _ _ _
What is the pH value of the 0.348 M Ba(OH)₂ solution? Let us continue the rest of the calculation.
Recall that [tex]\boxed{ \ pH + pOH = 14 \ }[/tex]
[tex]\boxed{ \ pH = 14 - pOH \ }[/tex]
[tex]\boxed{ \ pH = 14 - 0.157 \ }[/tex]
∴ [tex]\boxed{ \ pH = 13.843 \ }[/tex]
A pH value closes up to 14 indicates that the solution represents a very strong base. In the same way, if the pH value is very close to one, it means the solution is a very strong acid.
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