100 mL of 0.2 M H_2SO_4 is added to 100 m | vedclass

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(A) The reaction is: $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$.Number of millimoles of $H_2SO_4 = 100 \ mL 	imes 0.2 \ M = 20 \ mmol$.Number of

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Acidic

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(A) The reaction is: $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$.Number of millimoles of $H_2SO_4 = 100 \ mL 	imes 0.2 \ M = 20 \ mmol$.Number of millimoles of $NaOH = 100 \ mL 	imes 0.2 \ M = 20 \ mmol$.Since $1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NaOH$,$20 \ mmol$ of $H_2SO_4$ would require $40 \ mmol$ of $NaOH$.We only have $20 \ mmol$ of $NaOH$,so $H_2SO_4$ is in excess.Remaining millimoles of $H_2SO_4 = 20 - (20/2) = 10 \ mmol$.Since each $H_2SO_4$ molecule provides $2 \ H^+$ ions,the concentration of $H^+$ ions is $2 	imes 10 \ mmol = 20 \ mmol$.Total volume $= 100 \ mL + 100 \ mL = 200 \ mL$.$[H^+] = 20 \ mmol / 200 \ mL = 0.1 \ M$.Since $[H^+] > 0$,the solution is acidic.

$100 \ mL$ of $0.2 \ M \ H_2SO_4$ is added to $100 \ mL$ of $0.2 \ M \ NaOH$. The resulting solution will be

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