Question

4) 10.00 mL of sulfuric acid is neutralized in a titration using 18.54 mL of 0.100 M NaOH.
a) Write the neutralization equation for this reaction.
b) What is the concentration of the sulfuric acid?

Answer 1

Answer:

a) $H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

b) $M_{H_2SO_4}=0.0927M$

Explanation:

Hello!

In this case, for this acid-base reaction which also known as a neutralization because sulfuric acid is neutralized with sodium hydroxide, we can write the undergoing chemical reaction as shown below:

$H_2SO_4+NaOH\rightarrow Na_2SO_4+H_2O$

However, it needs to be balanced as two sodium atoms are yielded:

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

Next, since there is a 1:2 mole ratio between the acid and the base, at the equivalence point, at which the moles of acid and base are consumed, we write:

$2n_{H_2SO_4}=n_{NaOH}$

Which can also be written in terms of the given volumes and concentration of the base:

$2M_{H_2SO_4}V_{H_2SO_4}=M_{NaOH}V_{NaOH}$

In such a way, we solve for the concentration of sulfuric acid as shown below:

$M_{H_2SO_4}=\frac{M_{NaOH}V_{NaOH}}{2V_{H_2SO_4}} =\frac{18.54mL*0.100M}{2*10.00mL}\\\\ M_{H_2SO_4}=0.0927M$

Best regards!