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Calculate the pH of 10.0mL deionized water plus 5.0mL of 0.10M NaOH
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Because NaOH is a strong base, you dont need to write an equilibrium expression; that is \([NaOH]=[OH^-]\)). Find the molarity (of sodium hydroxide) of the solution, then use it to find pOH and subsequently pH. \(Molarity=\dfrac{moles~of~NaOH}{L_{solution}}\) (you can also simply use the dilution formula for this step: \(M_1V_1=M_2V_2\)) Next use \(pOH=-log[OH^-]\) then pH+pOH=14 to find pH.
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