Answer:
THE MOLARITY OF THE ACID IS 0.232M
Explanation:
Molarity is the number of moles of solute per liter of a solution.
In titration, the molarity of the acid and the molarity of the base are related by this equation:
Ca VA / CbVb = na / nb
Ca = concentration of the acid = ?
Va = volume of the acid = 49 mL
Cb = concentration of the base = 0.333M
Vb = volume of the base = 68.4mL
Na = number of mole of acid = 1
Nb = number of mole of base = 2
Equation for the reaction:
H2SO4 + 2NaOH --------> Na2SO4 + 2H2O
Solving for the molarity of the acid, we have :
Ca = CbVbNa / VaNb
Ca = 0.333M * 68.4mL * 1 / 49mL *2
Ca = 22.7772 * 10^-3 / 98 *10^-3
Ca = 0.232M
The molarity of the acid is therefore 0.232M.
Answer:
[tex]M_{acid}=0.232M[/tex]
Explanation:
Hello,
In this case, since the reaction between sulfuric acid and sodium hydroxide is:
H2SO4+2NaOH⇒Na2SO4+2H2O
We find a 1:2 molar ratio between the acid and the base, for that reason, at neutralization we verify:
[tex]2\times n_{acid}=n_{base}[/tex]
That in terms of concentration and volumes:
[tex]2\times M_{acid}V_{acid}=M_{base}V_{base}[/tex]
With it, we can compute the molarity of the acid solution:
[tex]M_{acid}=\frac{ M_{base}V_{base}}{2\times V_{acid}}=\frac{0.333M*68.4mL}{2\times 49.0mL} \\\\M_{acid}=0.232M[/tex]
Best regards.
