# Question 9ef0b

Aug 23, 2016

M = .085 ${V}_{1} \times {M}_{1} = {V}_{2} \times {M}_{2}$

#### Explanation:

The volume times the molarity of acid $\left\{{H}^{+}\right\}$ equals the volume times the molarity of base $\left\{O {H}^{-}\right\}$

Li OH is a mono hydroxide base. so the molarity of LiOH equals the molarity of the base $\left\{O {H}^{-}\right)$

${H}_{2} C {O}_{3}$ is a bi hydrogen acid. so the molarity of the acid ${H}^{+}$
equals $2 \times M$ the molarity of ${H}_{2} C {O}_{3}$

${V}_{1} \times M \left(H +\right) = {V}_{2} \times M \left(O {H}^{-}\right)$

$47 m l \times 2 M \left({H}^{+}\right) = 37.5 m l \times 0.215$

M = $37.5 \times \frac{0.215}{47.0 \times 2}$

M = .0858

Aug 23, 2016

sf(0.085color(white)(l)"mol/l"#

#### Explanation:

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

This tells us that 1 mole of $\textsf{{H}_{2} C {O}_{3}}$ will be neutralised by 2 moles of $\textsf{L i O H}$.

To find the number of moles of $\textsf{L i O H}$ we can say that:

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

$\therefore$$\textsf{{n}_{L i O H} = c \times v = 0.215 \times \frac{37.5}{1000} = 0.008}$

From the equation we can say that:

$\textsf{{n}_{{H}_{2} C {O}_{3}} = \frac{0.008}{2} = 0.004}$

$\therefore$$\textsf{{c}_{{H}_{2} C {O}_{3}} = {n}_{{H}_{2} C {O}_{3}} / v = \frac{0.004}{\frac{47.0}{1000}} = 0.085 \textcolor{w h i t e}{l} \text{mol/l}}$