Respuesta :
Ba(OH)2 dissociates according to the equation below to yield Barium ions and hydroxide ions.
Ba(OH)2 = Ba²⁺ + 2 OH⁻
The concentration of Ba²⁺ is 1.0 ×10^-3 M
Thus that of OH⁻ ions will be 2× 1.0 ×10^-3 = 2.0 × 10^-3 M
Thus; the answer is 2.0 × 10^-3 M
Ba(OH)2 = Ba²⁺ + 2 OH⁻
The concentration of Ba²⁺ is 1.0 ×10^-3 M
Thus that of OH⁻ ions will be 2× 1.0 ×10^-3 = 2.0 × 10^-3 M
Thus; the answer is 2.0 × 10^-3 M
The [tex]{\text{O}}{{\text{H}}^ - }[/tex] concentration in a [tex]1.0 \times {10^{ - 3}}{\text{ M Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}[/tex] solution is [tex]\boxed{2.0 \times {{10}^{ - 3}}{\text{ M}}}[/tex].
Further Explanation:
Concentration is used to describe relationship between various components of solution. For this purpose, a wide variety of concentration terms are used. Some of these are enlisted below.
1. Molarity (M)
2. Mole fraction (X)
3. Molality (m)
4. Parts per million (ppm)
5. Mass percent ((w/w) %)
6. Volume percent ((v/v) %)
7. Parts per billion (ppb)
Molarity is defined as moles of solute present in one litre of solution. It is represented by M and its unit is mol/L. The expression for molarity of solution is as follows:
[tex]{\text{Molarity of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Volume }}\left( {\text{L}} \right){\text{ of solution}}}}[/tex]
[tex]{\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}[/tex] is a strong base and it dissociates into its respective ions as follows:
[tex]{\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}} \rightleftharpoons {\text{B}}{{\text{a}}^{2 + }} + 2{\text{O}}{{\text{H}}^ - }[/tex]
This indicates one mole of [tex]{\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}[/tex] dissociates to produce one mole of [tex]{\text{B}}{{\text{a}}^{2 + }}[/tex] ions and two moles of [tex]{\text{O}}{{\text{H}}^ - }[/tex] ions.
Given information:
[tex]{\text{Concentration of Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}} = 1.0 \times {10^{ - 3}}{\text{ M}}[/tex]
Since one mole of [tex]{\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}[/tex] forms two moles of [tex]{\text{O}}{{\text{H}}^ - }[/tex] ions, concentration of [tex]{\text{O}}{{\text{H}}^ - }[/tex] ions becomes twice the concentration of [tex]{\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}[/tex] and is calculated as follows:
[tex]\begin{aligned}{\text{Concentration of O}}{{\text{H}}^ - } &= 2\left( {1.0 \times {{10}^{ - 3}}{\text{ M}}} \right) \\&= 2.0 \times {10^{ - 3}}{\text{ M}}\\\end{aligned}[/tex]
Therefore concentration of [tex]{\text{O}}{{\text{H}}^ - }[/tex] ions comes out to be [tex]2.0 \times {10^{ - 3}}{\text{ M}}[/tex].
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Concentration terms
Keywords: concentration, concentration terms, solutions, molarity, molality, Ba(OH)2, OH-, Ba2+, 1.0*10^-3 M, 2.0*10^-3 M, molarity of solution, moles of solute, volume of solution, mole fraction, parts per million, parts per billion, mass percent, volume percent.
