# What volume of 0.181 M Na_3PO_4 solution is necessary to completely react with 91.0 mL of 0.107 M CuCl_2?

Mar 10, 2016

$35.9 \text{ml}$

#### Explanation:

$3 C u C {l}_{2 \left(a q\right)} + 2 N {a}_{3} P {O}_{4 \left(a q\right)} \rightarrow C {u}_{3} {\left(P {O}_{4}\right)}_{2 \left(s\right)} + 6 N a C l \left(a q\right)$

This tells us that 3 moles of $C u C {l}_{2}$ react with 2 moles $N {a}_{3} P {O}_{4}$.

$\therefore$ 1 mole $C u C {l}_{2}$ will react with 2/3 moles $N {a}_{3} P {O}_{4}$

We know that concentration = moles/volume i.e:

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

$\therefore n = c \times v$

$\therefore n C u C {l}_{2} = 0.107 \times \frac{91.0}{1000} = 9.737 \times {10}^{- 3}$

I divided by 1000 to convert $\text{ml}$ to $\text{L}$

$\therefore n N {a}_{3} P {O}_{4} = 9.737 \times {10}^{- 3} \times \frac{2}{3} = 6.491 \times {10}^{- 3}$

$v = \frac{n}{c} = \frac{6.491 \times {10}^{- 3}}{0.181} = 35.86 \times {10}^{- 3} \text{L}$

$\therefore v = 35.86 \text{ml}$