Question

How many moles of NaOH is needed to neutralize 45.0 ml of 0.30M H2SeO4? Question 2 options: A) 0.00675 B) 27.0 C) 0.027 D) 0.0135

Answer 1

C) 0.027

Step-by-step explanation:

In this case we can start with the reaction between
`NaOH` and
`H_2SeO_4`, so:

`H_2SeO_4~+~NaOH~->~Na_2SeO_4~+~H_2O`

We have an acid (
`H_2SeO_4`) and a base (
`NaOH`), therefore we will have an acid-base reaction in which a salt is produced (
`Na_2SeO_4`) and water (
`H_2O`).

Now we can balance the reaction:

`H_2SeO_4~+~2NaOH~->~Na_2SeO_4~+~2H_2O`

If we have the volume (45 mL= 0.045 L) and the concentration (0.3 M) of the acid we can calculate the moles using the molarity equation:

`M=(mol)/(L)`

`0.3~M~=~(mol)/(0.045~L)`

`mol=0.3~M*0.045~L=0.0135~mol~of~H_2SeO_4`

In the balanced reaction, we have a 2:1 molar ratio between the acid and the base (for each mol of
`H_2SeO_4` 2 moles of
`NaOH` are consumed), with this in mind we can calculate the moles of NaOH:

`0.0135~mol~of~H_2SeO_4(2~mol~NaOH)/(1~mol~of~H_2SeO_4)=0.027~mol~NaOH`

I hope it helps!