Given that –
We know that –
Then –
[tex]\qquad[/tex][tex]\bf\longrightarrow pOH = 14 -12.5 [/tex]
[tex]\qquad[/tex][tex]\sf \longrightarrow pOH = 1.5[/tex]
[tex]\qquad[/tex][tex]\sf \longrightarrow [OH^- ] = 10^{-1.5 }[/tex]
[tex]\qquad[/tex][tex]\bf \longrightarrow [OH^-] = 3.16 × 10^{-2}[/tex]
Now the number of moles of KOH need to ensure that concentration of Hydroxide anions is equal to –
[tex]\qquad[/tex][tex]\bf \longrightarrow 3.74\: L \times \dfrac{3.16× 10^{-2}}{1 \: L }[/tex]
[tex]\qquad[/tex][tex]\bf \longrightarrow 1. 18 × 10^{-1 }M [/tex]
Volume of the solution contains the need number of moles of Hydroxide anions –
[tex]\qquad[/tex][tex]\sf \longrightarrow \dfrac{ 1.18×10^{-1} \: moles \: OH^-}{0.826 \: moles \: OH^-}[/tex]
[tex]\qquad[/tex][tex]\sf \longrightarrow 0.143 L [/tex]
[tex]\qquad[/tex][tex]\sf \longrightarrow 0.143 \times 1000[/tex]
[tex]\qquad[/tex][tex]\pink{\bf\longrightarrow 143 mL }[/tex]
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