Answer:
1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.
Explanation:
[tex]Moles (n)=Molarity(M)\times Volume (L)[/tex]
Moles of hydrochloric acid = n
Volume of hydrochloric acid solution = 200.0 mL = 0.200 L
Molarity of the hydrochloric acid = 0.089 M
[tex]n=0.089 M\times 0.200 L=0.0178 mol[/tex] of HCL
[tex]HCl(aq)+NaHCO_3(aq)\rightarrow NaCl(aq)+H_2O(l)+CO_2(g)[/tex]
According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.
Then 0.0178 moles of HCl wil be neutralized by :
[tex]\frac{1}{1}\times 0.0178 mol=0.0178 mol[/tex] of sodium bicarbonate
Mass of 0.0178 moles of sodium bicarbonate:
0.0178 mol × 72 g/mol = 1.4952 g
1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.