(B) The molar mass of $NaOH$ is $40 \, g/mol$.The number of moles of $NaOH$ is $n = \frac{4.0 \, g}{40 \, g/mol} = 0.1 \, mol$.Since the volume of the

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(B) The molar mass of $NaOH$ is $40 \, g/mol$.The number of moles of $NaOH$ is $n = \frac{4.0 \, g}{40 \, g/mol} = 0.1 \, mol$.Since the volume of the solution is $1 \, L$,the molarity $[NaOH] = 0.1 \, M$.$NaOH$ is a strong base,so $[OH^-] = 0.1 \, M = 10^{-1} \, M$.$pOH = -\log[OH^-] = -\log(10^{-1}) = 1$.Since $pH + pOH = 14$,we have $pH = 14 - 1 = 13$.

Class 11 Chemistry 2 Equilibrium: Ionic Equilibrium pH of strong Acids and strong Bases

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If $4.0 \, g$ of $NaOH$ is present in $1 \, L$ of solution,then its $pH$ will be:

A
$6$
B
$13$
C
$18$
D
$24$

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2 Equilibrium: Ionic Equilibrium

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pH of strong Acids and strong Bases

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If $4.0 \, g$ of $NaOH$ is present in $1 \, L$ of solution,then its $pH$ will be:

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