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

Answer:

C) 0.027

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:

$M=\frac{mol}{L}$

$0.3~M~=~\frac{mol}{0.045~L}$

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

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:

$0.0135~mol~of~H_2SeO_4\frac{2~mol~NaOH}{1~mol~of~H_2SeO_4}=0.027~mol~NaOH$

I hope it helps!