Question

$1.00$ L of $0.15M$ $NaOH$ absorbed $11.2$ millimoles of $C{O}_2$ from air. Hence, new molarity of $NaOH$ is:

• A. $0.1276$ M
• B. $0.1500$ M
• C. $0.0224$ M
• D. $0.0112$ M

Answer 1

Answer: (A)

$0.1276$ M

The reaction is as follows:
$2NaOH\left(2mol\right)+C{O}_2\left(1mol\right)\to N{a}_{2}C{O}_3\left(1mol\right)+H_{2}O$
$1$ mol $C{O}_2$ reacts with $2$ mol $NaOH$.
$\therefore 11.2\times {10}^{-3}molC{O}_2$ reacts with $0.0224$ mol of $NaOH$.
$NaOH$ at start $=0.15$ mol.
$NaOH$ unreacted $=0.150-0.224=0.1276$ in $1$ L.