10.0 , mL of Na_{2}CO_{3} solution is titrate | vedclass

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Question

(C) The most consistent and precise volume of $HCl$ obtained from the readings is $5.0 \, mL$.At the equivalence point,the number of milliequivalents

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(C) The most consistent and precise volume of $HCl$ obtained from the readings is $5.0 \, mL$.At the equivalence point,the number of milliequivalents $(meq)$ of $Na_{2}CO_{3}$ equals the number of milliequivalents of $HCl$.Let the molarity of $Na_{2}CO_{3}$ solution be $M$.The reaction is: $Na_{2}CO_{3} + 2HCl ightarrow 2NaCl + H_{2}O + CO_{2}$.The n-factor for $Na_{2}CO_{3}$ is $2$ and for $HCl$ is $1$.Using the formula: $M_{1} 	imes V_{1} 	imes n_{1} = M_{2} 	imes V_{2} 	imes n_{2}$$M 	imes 10 \, mL 	imes 2 = 0.2 \, M 	imes 5.0 \, mL 	imes 1$$20 	imes M = 1.0$$M = 0.05 \, mol/L$To convert to $mM$ (millimolar),we multiply by $1000$:$0.05 	imes 1000 = 50 \, mM$.

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