Question

To neutralize 100 g of HNO3 solution, 30 cm3 of 15% NaOH solution with a density of 1.15 g / cm3 was used. Calculate the mass fraction (in%) of nitric acid in the solution!

Answer 1

Let the mass of the total solution =100 gm

So mass of the $NaOH$ becomes=15 gm {That is 15% of the mass of the solution}

$d_{solution}=\dfrac{m_{solution}}{V_{solution}}$

$V_{solution}=\dfrac{100}{1.15} =86.95ml$

$n_{NaOH}=\dfrac{15}{40}=0.375moles$

Molarity of NaOH $=$ $\dfrac{n_{NaOH}}{V_{solution}}\times1000=\dfrac{0.375}{86.95}\times1000=4.31M$

Moles of NaOH required $=4.31\times \dfrac{30}{1000}=0.129 moles$

Moles of $HNO_3$ needed $(n_{HNO_3})$ $=0.129moles$

$n_{HNO_3}=\dfrac{m_{HNO_3}}{molarmass}$

Mass of $HNO_3=0.129\times63=8.127gm$

Now mass fraction of $HNO_3=\dfrac{8.127}{100}\times100=8.127$ %