Respuesta :
Explanation:
First, we will calculate the number of moles of both HCl and NaOH as follows.
No. of moles of HCl = Molarity × Volume
= [tex]1.2 M \times 0.01 L[/tex] (as 1 L = 1000 ml)
= 0.012 mol
No. of moles of NaOH = Molarity × Volume
= [tex]0.50 \times 0.1245 L[/tex] (12.45 ml = 0.1245 L)
= 0.006 mol
Therefore, the amount of HCl reacts with [tex]NaHCO_{3}[/tex] will be as follows.
(0.012 - 0.06) mol
= 0.006 mol
Reaction equation for this reaction is as follows.
[tex]NaHCO_{3} + HCl \rightarrow HCO^{-}_{3} + NaCl[/tex]
Here, one mole of HCl reacts with one mole of [tex]NaHCO_{3}[/tex] and it forms one mole of [tex]HCO^{-}_{3}[/tex]. Hence, moles of [tex]HCO^{-}_{3}[/tex] is 0.006 mol.
So, [tex]HCO^{-}_{3} + H_{2}O \rightarrow H_{2}CO_{3} + OH^{-}[/tex]
Here also, i mole of [tex]HCO^{-}_{3}[/tex] gives one mole of [tex]OH^{-}[/tex].
So, [tex]OH^{-}[/tex] = 0.006 moles
Now, moles of [tex]OH^{-}[/tex] present per gram will be calculated as follows.
Moles of [tex]OH^{-}[/tex] per gram = [tex]\frac{0.006 mol}{1.25 g}[/tex]
= 0.0048 g/mol
Thus, we can conclude that the moles of antacid activity per gram of baking soda is 0.0048 g/mol.
