Question

What is the molarity of a solution of sulfuric acid if 49.0 mL of it are neutralized by a titration

Answer 1

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 2

Answer:

$M_{acid}=0.232M$

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:

$2\times n_{acid}=n_{base}$

That in terms of concentration and volumes:

$2\times M_{acid}V_{acid}=M_{base}V_{base}$

With it, we can compute the molarity of the acid solution:

$M_{acid}=\frac{ M_{base}V_{base}}{2\times V_{acid}}=\frac{0.333M*68.4mL}{2\times 49.0mL} \\\\M_{acid}=0.232M$

Best regards.