pH= 11.332
KOH⇒K⁺+OH⁻
KOH(MW=56,1056 g/mol)
mol KOH :
[tex]\tt \dfrac{0.0256}{56.1056}=4.56\times 10^{-4}[/tex]
molarity KOH :
[tex]\tt \dfrac{4.56\times 10^{-4}}{0.212}=2.15\times 10^{-3}[/tex]
[OH⁻]=[KOH]=2.15 x 10⁻³
pOH=-log[OH⁻]
[tex]\tt pOH=-log~2.15\times 10^{-3}\\\\pOH=3-log~2.15\\\\pOH=2.668[/tex]
pH+pOH=14
[tex]\tt pH=14-2.668=\boxed{\bold{11.332}}[/tex]