Question

Question #227e7

Answer 1

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)