Question

g If the titration of a 10.0-mL sample of sulfuric acid requires 28.15 mL of 0.100 M sodium hydroxide, what is the molarity of the acid

Answer 1

`M_(acid)=0.141M`

Step-by-step explanation:

Hello,

In this case, the reaction between sulfuric acid and hydroxide is:

`H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O`

We can notice a 1:2 molar ratio between the acid and the base respectively, therefore, at the equivalence point we have:

`2*n_(acid)=n_(base)`

And in terms of volumes and concentrations:

`2*M_(acid)V_(acid)=M_(base)V_(base)`

So we compute the molarity of sulfuric acid as shown below:

`M_(acid)=(M_(base)V_(base))/(2*V_(acid)) =(0.100M*28.15mL)/(2*10.0mL)\\ \\M_(acid)=0.141M`

Best regards.