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$20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}$ is added to the $30 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{NaOH}$. To this solution extra $50 \mathrm{~mL}$ of water was added. What is the molarity of the final solution formed?
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$0.01 \mathrm{M}$
$\mathrm{n}_{\mathrm{HCl}}=0.1 \mathrm{M} \times 0.02 \mathrm{~L}=0.002 \mathrm{~mol}$
$\mathrm{n}_{\mathrm{NaOH}}=0.1 \mathrm{M} \times 0.03 \mathrm{~L}=0.003 \mathrm{~mol}$
Number of moles of $\mathrm{NaOH}$ remaining after neutralization. $=0.003-0.002=0.001 \mathrm{~mol}$.
Total final volume $=50 \mathrm{~mL}+50 \mathrm{~mL}=100 \mathrm{~mL}=0.1 \mathrm{~L}$ $\Rightarrow$ Molarity $=\frac{0.001}{0.1}=0.01 \mathrm{M}$
$\mathrm{n}_{\mathrm{NaOH}}=0.1 \mathrm{M} \times 0.03 \mathrm{~L}=0.003 \mathrm{~mol}$
Number of moles of $\mathrm{NaOH}$ remaining after neutralization. $=0.003-0.002=0.001 \mathrm{~mol}$.
Total final volume $=50 \mathrm{~mL}+50 \mathrm{~mL}=100 \mathrm{~mL}=0.1 \mathrm{~L}$ $\Rightarrow$ Molarity $=\frac{0.001}{0.1}=0.01 \mathrm{M}$
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