Calculate the pH of the solution containing 2g of NaOH per liter of the solution.

Explanation

Molarity of NaOH
(M) `= ("Mass of NaOH")/("mol. mass") xx (1000)/("vol. of soluton (mL)")`
`= (2)/(40) xx (1000)/(1000) = 0.05 M`
Now, `NaOH overset(aq)rarr Na^(+)(aq) + OH^(-)(aq), 0.05 = 5 xx 10^(-2)M`
Now `[H^(+)] [OH^(-)] = 1 xx 10^(-14)`
`[H^(+)] = (1 xx 10^(-14))/([OH^(-)]) = (1 xx 10^(-14))/(5 xx 10^(-2)) = 2 xx 10^(-13)M`
pH = `-log[H^(+)] = -log (2 xx 10^(-13))`
`= - log 2 - log 10^(-13)`
`= -log 2 + 13 log 10`
`= -0.3010 + 13 = 12.6990`