If 0.4 g of NaOH is present in 1 L of solut | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The molar mass of $NaOH$ is $40 \ g/mol$.The number of moles of $NaOH = \frac{0.4 \ g}{40 \ g/mol} = 0.01 \ mol$.Since the volume of the solution

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) The molar mass of $NaOH$ is $40 \ g/mol$.The number of moles of $NaOH = \frac{0.4 \ g}{40 \ g/mol} = 0.01 \ mol$.Since the volume of the solution is $1 \ L$,the molarity $[NaOH] = 0.01 \ M = 10^{-2} \ M$.Since $NaOH$ is a strong base,$[OH^-] = [NaOH] = 10^{-2} \ M$.Using the ionic product of water,$[H^+][OH^-] = 10^{-14}$,we get $[H^+] = \frac{10^{-14}}{10^{-2}} = 10^{-12} \ M$.$pH = -\log[H^+] = -\log(10^{-12}) = 12$.

If $0.4 \ g$ of $NaOH$ is present in $1 \ L$ of solution,then its $pH$ will be:

Similar Questions

The highest $pH$ $(14)$ will be of which of the following?

The $pH$ of $100 \ mL$ of an aqueous solution of a strong acid is $1$. The volume of water to be added to change its $pH$ to $2$ is .......... $mL$.

What will be the concentration of $H^{+}$ ions in a solution prepared by dissolving $4 \ g$ of $NaOH$ in $100 \ mL$ of water at $25 \ ^{\circ}C$?

What is the $pH$ of the $NaOH$ solution when $0.04 \ g$ of it is dissolved in water to make a $100 \ mL$ solution?

The $pH$ of a completely dissociated $0.005 \ M \ H_2SO_4$ solution is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module