# Question #93f99

Nov 30, 2014

The molarity of the acid is 1.434.

$2 H N {O}_{3 \left(a q\right)} + C a {\left(O H\right)}_{2 \left(a q\right)} \rightarrow C a {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + 2 {H}_{2} {O}_{\left(l\right)}$

This tells us that 2 moles of $H N {O}_{3}$ reacts with 1 mole of $C a {\left(O H\right)}_{2}$

$c = \frac{n}{v}$
So $n = c v$

So no. moles $C a {\left(O H\right)}_{2} = \frac{17.42 \times 0.823}{1000} = 0.01434$

So no. moles $H N {O}_{3} = 0.01434 \times 2 = 0.02868$

$c = \frac{n}{v} = \frac{0.02868}{0.02} = 1.434 m o l . {\mathrm{dm}}^{- 3}$