# Question #227e7

Dec 14, 2014

23.36 ml is required.

${H}_{2} S {O}_{4 \left(a q\right)} + 2 N a O {H}_{\left(a q\right)} \rightarrow N {a}_{2} S {O}_{4 \left(a q\right)} + 2 {H}_{2} {O}_{\left(l\right)}$

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

So $n = c v$

So we can get the no. moles of ${H}_{2} S {O}_{4}$:

$n = 0.102 \times \frac{15}{1000} = 1.53 \times {10}^{- 3} m o l$

From the equation we can see that the no. moles $N a O H$ must be 2x this amount:

$n N a O H = 1.53 \times {10}^{- 3} \times 2 = 3.06 \times {10}^{- 3}$

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

So $v = \frac{n}{c} = \frac{3.06 \times {10}^{- 3}}{0.131} = 23.36 \times {10}^{- 3} {\mathrm{dm}}^{3}$

$v = 23.36 c {m}^{3}$

(or ml)