Sodium carbonate is a reagent that may be used to standardize acids in the same way you used KHP in this experiment. In such a standardization, it was found that a 0.512 g sample of sodium carbonate required 26.30 mL of a sulfuric acid solution to reach the end point for the reaction. Na2CO3 (aq)+ H2SO4 (aq) ---> H2O (l)+ CO2 (g) + Na2SO4 (aq)What is the molarity of H2SO4?

M = 0.1825 M

Explanation:

To do this, let's write the equation again:

Na₂CO₃ + H₂SO₄ > H₂O + CO₂ + Na₂SO₄

As we can see, the equation is already balanced and we can also see that the mole ratio between the acid and the carbonate is 1:1, this means that the moles of the acid, would be the same moles of the carbonate, therefore, we can use the following expression:

M₁V₁ = M₂V₂ (1)

1: Is the carbonate

2: is the acid

To get the concentration of the acid, we need to calculate the moles of the carbonate used. This can be done using the molecular mass of the sodium carbonate, which is 105.9888 g/mol, so the moles:

n₁ = 0.512 / 105.9888 = 0.0048 moles

Now that we have the moles, we can use (1) and calculate the concentration of the acid.

We know that:

n₁ = M₁V₁ (2)

Replacing in (1) we have:

n₁ = M₂V₂

M₂ = n₁ / V₂ (3)

Now all we have to do is replace the values and solve for the concentration:

M₂ = 0.0048 / (0.02630)

answer; natural resources;

your answer would be “dry”