Question

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Find the pOH of a 0.35 {~mol} / {LH}_{2} {SO}_{4} solution.

Answer 1

To find the pOH of a 0.35 mol/L
`\displaystyle\sf H_(2)SO_(4)` solution, we need to determine the hydroxide ion concentration (
`\displaystyle\sf [OH^(-)]`) and then calculate the pOH.

Since
`\displaystyle\sf H_(2)SO_(4)` is a strong acid, it completely dissociates in water. It produces two moles of H+ ions for every mole of
`\displaystyle\sf H_(2)SO_(4)`. Therefore, the concentration of H+ ions is also 0.35 mol/L.

In a neutral solution, the concentration of
`\displaystyle\sf OH^(-)` ions is equal to the concentration of H+ ions, which is 0.35 mol/L in this case.

To calculate pOH, we can use the formula:

`\displaystyle\sf pOH=-\log_(10) [OH^(-)]`

Substituting the value of
`\displaystyle\sf [OH^(-)]=0.35`, we have:

`\displaystyle\sf pOH=-\log_(10) (0.35)`

Calculating this using a calculator or logarithmic table, we find that the pOH of the 0.35 mol/L
`\displaystyle\sf H_(2)SO_(4)` solution is approximately 0.46.

Therefore, the pOH of the given solution is 0.46.