Blue litmus turns red in the following mixtur | vedclass

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

Question

(C) Blue litmus turns red if the mixture is acidic,i.e.,moles of $H^+ >$ moles of $OH^-$.$(A)$ Moles of $H^+ = (100 	imes 10^{-3} \ L) 	imes (2 	im

Vedclass · Accepted Answer

$100 \ mL$ of $1 	imes 10^{-2} \ M \ H_2SO_4 + 100 \ mL$ of $1 	imes 10^{-2} \ M \ NaOH$

Vedclass · Answer

(C) Blue litmus turns red if the mixture is acidic,i.e.,moles of $H^+ >$ moles of $OH^-$.$(A)$ Moles of $H^+ = (100 	imes 10^{-3} \ L) 	imes (2 	imes 10^{-2} \ mol \ L^{-1}) = 2 	imes 10^{-3} \ mol$.Moles of $OH^- = (100 	imes 10^{-3} \ L) 	imes (2 	imes 10^{-2} \ mol \ L^{-1}) = 2 	imes 10^{-3} \ mol$.Since moles of $H^+ = OH^-$,the solution is neutral.$(B)$ Moles of $H^+ = (100 	imes 10^{-3} \ L) 	imes (10^{-2} \ mol \ L^{-1}) = 10^{-3} \ mol$.Moles of $OH^- = (100 	imes 10^{-3} \ L) 	imes (2 	imes 10^{-2} \ mol \ L^{-1}) = 2 	imes 10^{-3} \ mol$.Since moles of $H^+ $(C)$ Moles of $H^+ = (100 	imes 10^{-3} \ L) 	imes (2 	imes 10^{-2} \ mol \ L^{-1}) = 2 	imes 10^{-3} \ mol$.Moles of $OH^- = (100 	imes 10^{-3} \ L) 	imes (10^{-2} \ mol \ L^{-1}) = 10^{-3} \ mol$.Since moles of $H^+ > OH^-$,the solution is acidic.$(D)$ Moles of $H^+ = (100 	imes 10^{-3} \ L) 	imes (10^{-2} \ mol \ L^{-1}) = 10^{-3} \ mol$.Moles of $OH^- = (100 	imes 10^{-3} \ L) 	imes (10^{-2} \ mol \ L^{-1}) = 10^{-3} \ mol$.Since moles of $H^+ = OH^-$,the solution is neutral.Therefore,option $(C)$ is correct.

Blue litmus turns red in the following mixture of acid and base:

Similar Questions

Calculate the ionic product of the solution formed by mixing equal volumes of $0.004 \, M \, CaSO_4$ and $0.002 \, M \, H_2SO_4$.

In a solution,$0.02 \ M$ acetic acid is $4 \%$ dissociated. The $[OH^{-}]$ in the solution is

Arrange the following $0.1 \ M$ salt solutions in increasing order of their $pH$:

The conductivity of a weak acid $HA$ of concentration $0.001 \, mol \, L^{-1}$ is $2.0 \times 10^{-5} \, S \, cm^{-1} .$ If $\Lambda_{m}^{\circ} (HA)=190 \, S \, cm^{2} \, mol^{-1}$,the ionization constant $(K_{a})$ of $HA$ is equal to $....\, \times 10^{-6} .$ (Round off to the Nearest Integer)

The molar conductivity of $0.01 \text{ M}$ monobasic acid at $25 \text{ }^\circ\text{C}$ is $15 \text{ ohm}^{-1} \text{cm}^2 \text{mol}^{-1}$. The molar conductivity of the same acid at infinite dilution is $375 \text{ ohm}^{-1} \text{cm}^2 \text{mol}^{-1}$. Calculate $[H^+]$ in solution.

Vedclass Test Series

Exam Paper Generator

Online Exam Module