Answer:
a. [tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
b. 0.092 M
Explanation:
Hello!
In this case, since for this titration process we know the used volume of 0.75-M sodium hydroxide needed to neutralize 155 mL of the acid, we first need to write the undergoing chemical reaction between them:
[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
This, since there is 1:2 mole ratio between the acid and the base, at the equivalent point we must respect:
[tex]2*n_{acid}=n_{base}[/tex]
That in terms of concentrations and volumes is:
[tex]2*M_{acid}V_{acid}=M_{base}V_{base}[/tex]
Thus, the concentration of acid was:
[tex]M_{acid}=\frac{M_{base}V_{base}}{2*V_{acid}} \\\\M_{acid}=\frac{0.75M*38mL}{2*155mL} \\\\M_{acid}=0.092M[/tex]
Best regards!
