Answer:
[tex][OH^-]=0.00086\frac{molOH}{L}[/tex]
[tex][H^+]=1.16x10^{-11}\frac{molH^+}{L}[/tex]
Explanation:
Hello,
Considering the dissociation chemical reaction:
[tex]Ba(OH)_2--->2OH^-+Ba^{+2}[/tex]
Based on the stoichiometry, one computes the concentration of the [tex]OH^-[/tex] via:
[tex][OH^-]=0.00043\frac{molBa(OH)_2}{L} *\frac{2molOH^-}{1molBa(OH)_2} =0.00086\frac{molOH}{L}[/tex]
Now, by using the following relationship including the water's dissociation constant, one computes the concentration of the [H^+]
[tex]Kw=[H^+][OH^-]\\H^+=\frac{Kw}{[OH^-]}=\frac{1x10^{-14}}{0.00086} =1.16x10^{-11}\frac{molH^+}{L}[/tex]
Best regards.
A 0.00043 M solution of Ba(OH)₂ has a concentration of OH⁻ of 8.6 × 10⁻⁴ M and a concentration of H⁺ of 1.2 × 10⁻¹¹ M.
What are the concentrations of OH⁻ and H⁺ in a 0.00043 M solution of Ba(OH)₂ at 25 °C? Assume complete dissociation.
Ba(OH)₂ is a strong base that dissociates according to the following equation.
Ba(OH)₂(aq) ⇒ Ba²⁺(aq) + 2 OH⁻(aq)
The molar ratio of Ba(OH)₂ to OH⁻ is 1:2. The concentration of OH⁻ in a 0.00043 M Ba(OH)₂ solution is:
[tex]\frac{0.00043molBa(OH)_2}{L} \times \frac{2molOH^{-} }{1molBa(OH)_2} = \frac{8.6 \times 10^{-4}molOH^{-} }{L}[/tex]
Then, we can calculate the concentration of H⁺ using the following expression.
[tex]Kw = 1.0 \times 10^{-14} = [H^{+} ][OH^{-} ]\\[H^{+} ] = \frac{1.0 \times 10^{-14}}{[OH^{-} ]} = \frac{1.0 \times 10^{-14}}{8.6 \times 10^{-4} } = 1.2 \times 10^{-11} M[/tex]
A 0.00043 M solution of Ba(OH)₂ has a concentration of OH⁻ of 8.6 × 10⁻⁴ M and a concentration of H⁺ of 1.2 × 10⁻¹¹ M.
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