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

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

b)
`M_(H_2SO_4)=0.0927M`

Step-by-step 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)=(M_(NaOH)V_(NaOH))/(2V_(H_2SO_4)) =(18.54mL*0.100M)/(2*10.00mL)\\\\ M_(H_2SO_4)=0.0927M`

Best regards!