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Question
Calculate the pH of the following solution:
0.3 g of NaOH dissolved in water to give 200 mL of solution.
Solution
For 0.3 g of NaOH dissolved in water to give 200 mL of solution:
\[\ce{NaOH -> Na^+_{(aq)} + OH^-_{(aq)}}\]
`["NaOH"] = 0.3 xx 1000/200 = 1.5 "M"`
`["OH"_("aq")^-] = 1.5 "M"`
Then `["H"^+] = 10^(-14)/1.5`
`= 6.66 xx 10^(-13)`
`"pH" = - log(6.66 xx 10^(-13))`
= 12.18
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