Blue litmus turns red in the following mixtur | vedclass

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(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 $1 	imes 10^{-2} \ M \ NaOH$

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(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:

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