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4.0 g of NaOH are dissolved per litre. Find (i) molarity of the solution (ii) OH^(-)ion concentration (iii) pH value of the solution (At. Masses : Na = 23, O=16, H=1).

Answer» <html><body><p></p>Solution :(i) Calculation of molarity : Mass of NaOH <a href="https://interviewquestions.tuteehub.com/tag/dissolved-956358" style="font-weight:bold;" target="_blank" title="Click to know more about DISSOLVED">DISSOLVED</a> = 4.0 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>/<a href="https://interviewquestions.tuteehub.com/tag/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a> <br/> Mol . Mass of NaOH = 40 `:.` Molarity of the solution `= ("Strength in g // litre")/("Mol . Mass") = (4.0)/(40) = 0.1 M` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) Calculation of the `OH^(-)` ion conc. <br/> NaOH completely ionizes as : `NaOH rarr Na^(+) + OH^(-)" " :. [OH^(-)]=[NaOH]=0.1 M = 10^(-1)M` <br/> (iii) Calculationof pH :<br/> We know that `[H_(3)O^(+)][OH^(-)]=K_(w)=1.0xx10^(-14)` <br/> `:. [H_(3)O^(+)]=(K_(w))/([OH^(-)])=(1.0xx10^(-14))/(10^(-1))=10^(-13)M" " :. pH = - <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a> [H_(3)O^(+)]= - log 10 ^(-13) = 13`.</body></html>


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