Answer:
a) [tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
b) [tex]M_{H_2SO_4}=0.0927M[/tex]
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:
[tex]H_2SO_4+NaOH\rightarrow Na_2SO_4+H_2O[/tex]
However, it needs to be balanced as two sodium atoms are yielded:
[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
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:
[tex]2n_{H_2SO_4}=n_{NaOH}[/tex]
Which can also be written in terms of the given volumes and concentration of the base:
[tex]2M_{H_2SO_4}V_{H_2SO_4}=M_{NaOH}V_{NaOH}[/tex]
In such a way, we solve for the concentration of sulfuric acid as shown below:
[tex]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[/tex]
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