40 mg of pure sodium hydroxide is dissolved in 10 L of distilled water. The pH of the solution is

Options

A. `9.0`
B. `10`
C. `11`
D. `12`

Answer

B

Explanation

`n_(NaOH)=("Mass"_(NaOH))/("Molar mass"_(NaOH))`
`=(40xx10^(-3)g)/(40g mol^(-1))`
`= 10^(-3)mol`
`"Molarity"_(NaOH)= (n_(NaOH))/(V_(L))= (10^(-3)mol)/(10L)`
Since NaOH is a strong monoacidic base,
`C_(OH^(-))=C_(NaOH)=10^(-4)mol L^(-1)`
`:. pOH=4(C_(OH^(-))=10^(-pOH)mol L^(-1))`
For every aqueous solution at 298 K, we have
`pH+pOH=14`
or `pH=14-pOH`
`=14-4`
`=10`