Sodium carbonate is a reagent that may be used to standardize acids in the same way you used πΎπ»π in this experiment. In such a standardization, it was found that a 0.512 π sample of sodium carbonate required 26.30 ππΏ of a sulfuric acid solution to reach the end point for the reaction. ππ2πΆπ3 (ππ) + π»2ππ4 (ππ) β π»2π (π) + πΆπ2 (π) + ππ2ππ4(ππ) What is the molarity of the π»2ππ4?Β
Answer:
M = 0.1825 M
Explanation:
To do this, let's write the equation again:
NaβCOβ + HβSOβ ---------> HβO + COβ + NaβSOβ
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:
MβVβ = MβVβ (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:
nβ = 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:
nβ = MβVβ (2)
Replacing in (1) we have:
nβ = MβVβ
Mβ = nβ / Vβ (3)
Now all we have to do is replace the values and solve for the concentration:
Mβ = 0.0048 / (0.02630)
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