Answer:
3 acidic hydrogens per molecule of citric acid
Explanation:
In a sample of 37.2 mL(0.0372 L) of 0.105 mol/L of NaOH, will have:
n = 0.105x0.0372 = 0.0039 mol of NaOH
The dissociation of NaOH will give the same number of moles of Na⁺ and OH⁻.
The molar mass of citric acid is:
C: 12g/mol x 6 = 72 g/mol
H: 1g/mol x 8 = 8g/mol
O: 16 g/mol x 7 = 112 g/mol
192 g/mol
So, 0.250g of the acid has
n = mass/molar mass
n = 0.250/192
n = 0.0013 mol.
To be neutralized, it will be necessary 0.0039 mol of acidic hydrogens to react with the 0.0039 mol of OH⁻.
The dissociation reaction of one molecule of the acid will give the stoichiometry:
1 mol of acid ----------------------- x mol of acidic hydrogens
0.0013 mol --------------------------- 0.0039
For a simple direct three rule:
0.0013x = 0.0039
x = 3 acidic hydrogens per molecule of citric acid.
