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What is the [tex]\([OH^-]\)[/tex] in a solution with a pOH of 4.22?

A. [tex]\(1.7 \times 10^{-10} M\)[/tex]

B. [tex]\(6.0 \times 10^{-5} M\)[/tex]

C. [tex]\(6.3 \times 10^{-1} M\)[/tex]

D. [tex]\(1.7 \times 10^4 M\)[/tex]


Sagot :

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].