Answer :
To find the [tex]\([ \text{OH}^- ]\)[/tex] concentration in a solution with a given pOH, we can use the relationship between pOH and hydroxide ion concentration. The pOH is defined as the negative logarithm (base 10) of the hydroxide ion concentration:
[tex]\[ \text{pOH} = -\log [ \text{OH}^- ] \][/tex]
Given the pOH, we can rearrange this equation to solve for [tex]\([ \text{OH}^- ]\)[/tex]:
[tex]\[ [ \text{OH}^- ] = 10^{-\text{pOH}} \][/tex]
In this problem, the pOH of the solution is given as 4.22. Plugging this value into the equation:
[tex]\[ [ \text{OH}^- ] = 10^{-4.22} \][/tex]
Calculating this value, we get:
[tex]\[ [ \text{OH}^- ] \approx 6.0255958607435806 \times 10^{-5} \, \text{M} \][/tex]
Therefore, the closest answer that matches this concentration is:
B. [tex]\( \quad 6.0 \times 10^{-5} \, \text{M} \)[/tex]
This solution verifies that the hydroxide ion concentration in a solution with a pOH of 4.22 is approximately [tex]\( 6.0 \times 10^{-5} \, \text{M} \)[/tex].
[tex]\[ \text{pOH} = -\log [ \text{OH}^- ] \][/tex]
Given the pOH, we can rearrange this equation to solve for [tex]\([ \text{OH}^- ]\)[/tex]:
[tex]\[ [ \text{OH}^- ] = 10^{-\text{pOH}} \][/tex]
In this problem, the pOH of the solution is given as 4.22. Plugging this value into the equation:
[tex]\[ [ \text{OH}^- ] = 10^{-4.22} \][/tex]
Calculating this value, we get:
[tex]\[ [ \text{OH}^- ] \approx 6.0255958607435806 \times 10^{-5} \, \text{M} \][/tex]
Therefore, the closest answer that matches this concentration is:
B. [tex]\( \quad 6.0 \times 10^{-5} \, \text{M} \)[/tex]
This solution verifies that the hydroxide ion concentration in a solution with a pOH of 4.22 is approximately [tex]\( 6.0 \times 10^{-5} \, \text{M} \)[/tex].
