Question #227e7

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Answer 1 · 2014-12-14T00:54:47

23.36 ml is required.

$H_2SO_(4(aq))+2NaOH_((aq))rarrNa_2SO_(4(aq))+2H_2O_((l))$

$c=n/v$

So $n = cv$

So we can get the no. moles of $H_2SO_4$ :

$n = 0.102xx(15)/(1000)=1.53xx10^(-3)mol$

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

$nNaOH = 1.53xx10^(-3)xx2=3.06xx10^(-3)$

$c=n/v$

So $v=n/c=(3.06xx10^(-3))/(0.131)=23.36xx10^(-3)dm^3$

$v=23.36cm^3$

(or ml)