The volume of 0.04 M aqueous solution of HNO | vedclass

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(C) The neutralization reaction is: $2HNO_3 + Ba(OH)_2 \rightarrow Ba(NO_3)_2 + 2H_2O$. According to the law of equivalence,the number of equivalents

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) The neutralization reaction is: $2HNO_3 + Ba(OH)_2 ightarrow Ba(NO_3)_2 + 2H_2O$. According to the law of equivalence,the number of equivalents of $HNO_3$ must equal the number of equivalents of $Ba(OH)_2$. $n_{factor} 	imes M_1 	imes V_1 = n_{factor} 	imes M_2 	imes V_2$. For $HNO_3$,$n_{factor} = 1$. For $Ba(OH)_2$,$n_{factor} = 2$. $1 	imes 0.04 	imes V_1 = 2 	imes 0.01 	imes 50$. $0.04 	imes V_1 = 1$. $V_1 = \frac{1}{0.04} = 25 \ mL$.

The volume of $0.04 \ M$ aqueous solution of $HNO_3$ required to neutralize $50 \ mL$ of $0.01 \ M$ aqueous solution of $Ba(OH)_2$ is (Assume complete neutralization). (in $mL$)

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