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

Answer:

$M_{acid}=0.141M$

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:

$M_{acid}=\frac{M_{base}V_{base}}{2*V_{acid}} =\frac{0.100M*28.15mL}{2*10.0mL}\\ \\M_{acid}=0.141M$

Best regards.