Question

If 25.0 mL of an aqueous solution of H₂SO4 requires19.7 mL of 0.72 M NaOH to reach the end point, whatis the molarity of the H₂SO solution?

Answer 1

Answer: the molarity of the H2SO4 solution is 0.28 M

Step-by-step explanation:

The question requires us to calculate the molarity of a sulfuric acid (H2SO4) solution, knowing that it is necessary 19.7 mL of a 0.72 M sodium hydroxide (NaOH) solution to neutralize 25.0 mL of the acid solution.

To solve this problem, we'll need to consider the chemical reaction between H2SO4 and NaOH:

`H_2SO_(4(aq))+2NaOH_((aq))\rightarrow Na_2SO_(4(aq))+2H_2O_((l))`

Note that the chemical equation above is already balanced and it tells us that 2 moles of NaOH are necessary to completely react (in other words, to completely neutralize) 1 mol of H2SO4.

Therefore, at the end point of the titulation (where the neutralization occurs), we would need:

`2* n_(H_2SO_4)=1* n_(NaOH)`

where n corresponds to the number of moles of a substance.

Knowing that the molarity of a solution (C, in mol/L or M) is defined as the number of moles (n) divided by the volume of the solution (V, in liters), we can write the number of moles as the product between C and V:

`C=(n)/(V)\rightarrow n=C* V`

And we can rewrite the equation for the end point of the reaction as:

`2(C_(H_2SO_4)* V_(H_2SO_4))=1(C_(NaOH)* V_(NaOH))`

We can rearrange the equation above to calculate the molarity of H2SO4:

`C_(H_2SO_4)=(C_(NaOH)* V_(NaOH))/(2* V_(H_2SO_4))`

The question provided the following values:

C(H2SO4) = ? (needs to be calculated)

V(H2SO4) = 25.0mL

C(NaOH) = 0.72 M

V(NaOH) = 19.7mL

Replacing these values to calculate the molarity of H2SO4, we'll have:

`C_(H_2SO_4)=(0.72M*19.7mL)/(2*25.0mL)=0.28M`

Therefore, the molarity of the H2SO4 solution is 0.28 M.