Question #227e7

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Answer 1 · 2014-12-14T00:54:47

23.36 ml is required.

$H_{2} S O_{4 (a q)} + 2 N a O H_{(a q)} \to N a_{2} S O_{4 (a q)} + 2 H_{2} O_{(l)}$ $H_2SO_(4(aq))+2NaOH_((aq))rarrNa_2SO_(4(aq))+2H_2O_((l))$

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

So $n = c v$ $n = cv$

So we can get the no. moles of $H_{2} S O_{4}$ $H_2SO_4$ :

$n = 0.102 \times \frac{15}{1000} = 1.53 \times 10^{- 3} m o l$ $n = 0.102xx(15)/(1000)=1.53xx10^(-3)mol$

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

$n N a O H = 1.53 \times 10^{- 3} \times 2 = 3.06 \times 10^{- 3}$ $nNaOH = 1.53xx10^(-3)xx2=3.06xx10^(-3)$

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

So $v = \frac{n}{c} = \frac{3.06 \times 10^{- 3}}{0.131} = 23.36 \times 10^{- 3} d m^{3}$ $v=n/c=(3.06xx10^(-3))/(0.131)=23.36xx10^(-3)dm^3$

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

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