Question

A student performed a single titration using $2.00M$ $HCl$ to completely titrate $40.00$ mL of $1.00M$ $NaOH$. If the initial reading on the buret containing $HCl$ was $2.05$ mL, what will be the final reading?

• A. $82.05$ mL
• B. $42.05$ mL
• C. $20.00$ mL
• D. $10.00$ mL
• E. $22.05$ mL

Answer 1

The correct option is E $22.05$ mL

$M1V1=M2V2$
$M1=1.00M=$ molarity of $NaOH$
$V1=40.00mL=$ volume of $NaOH$
$M2=2.00M=$ molarity of $HCl$
$V2=$ Volume of $HCl$
$1.00\times 40=2.00\times V2$
$V2=20mL$
The initial burette reading is $2.05$ mL. The final burette reading will be
$2.05+20=22.05mL$