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Rama09 [41]
3 years ago
9

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

Chemistry
2 answers:
vaieri [72.5K]3 years ago
8 0

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.

Vedmedyk [2.9K]3 years ago
4 0

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.

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