Respuesta :
If 250 g of an unknown diprotic acid, and uses 30ml of .1m naoh to reach the second equivalence point in titrating this acid, then the molecular weight of this acid is 166.67 g/ mol.
Firstly, we will calculate the total number of moles of NaOH.
As we know that,
Molarity = Moles / volume
Moles = M × volume
Given,
Molarity = 0.1 M NaOH
volume = 0.03 L
By substituting all the values, we get
0.1 × 0.03 = moles
moles = 0.003 mol
As we know that the given acid is diprotic acid which means that the second equivalent point in titrating this acid reached, two moles of NaOH is treated by one moles of acid. Due to which the relationship is given as below:
0.003 × 1 mol acid / 2 mol NaOH = 0.0015 mole
Now, we will calculate the molar mass of diprotic acid as follow:
Molar mass = g acid /mol acid
= 0.250 g/ 0.0015 mol
= 166.67 g / mol.
Thus, we concluded that 250 g of an unknown diprotic acid, and uses 30ml of .1m naoh to reach the second equivalence point in titrating this acid, then the molecular weight of this acid is 166.67 g/ mol.
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