Question

Exactly 17.0 mL of a H2SO4 solution was required to neutralize 45.0 mL of 0.235 M NaOH. What was the concentration of the H2SO4 solution?

Answer 1

I get `0.31 \ "M"`

Explanation:

We have the balanced chemical equation:

`H_2SO_4(aq)+2NaOH(aq)->K_2SO_4(aq)+2H_2O(l)`

We first find out the moles of the base used, which is sodium hydroxide in this case.

We would use the formula:

`c=n/v`

• `n` is the number of moles

• `v` is the volume of solution

And so,

`n=c*v`

`=0.235 \ "M"*0.045 \ "L"`

`=(0.235 \ "mol")/(color(red)cancelcolor(black)"L")*0.045color(red)cancelcolor(black)"L"`

`=0.010575 \ "mol"`

So, we got `0.010575` moles of base. From the equation, one mole of acid neutralizes two moles of base, so here, we would need:

`(0.010575 \ "mol")/2=0.0052875 \ "mol"`

Now, we find the concentration of the solution of acid. And so,

`c=n/v=(0.0052875 \ "mol")/(0.017 \ "L")`

`~~0.31 \ "mol/L"`

`=0.31 \ "M"`