Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate.
`NaHCO_(3_((aq)))+HCl_((aq)) to NaCl_((eq))+H_(2)O_((l))+CO_(2_(g))`
How many milliliters of 0.125 M `NaHCO_(3)` solution are needed to neutralize 18.0 mL of 0.100 M HCI?

Options

A. `14.4 ` mL
B. 12.0 mL
C. 14.0 mL
D. 13.2 mL

Answer

A

Explanation

Given `M_(HCl)=0.1m, V_(HCl)=18.0mL`
`M_(NaHCO_(3))=0.125M, V_(NaHCO_(3))=?`
On applying `M_(HCl)xxV_(HCl)=M__(NaHCO_(3))xxV_(NaHCO_(3))`
`Rightarrow 0.1xx18=0.125xxF_(NaHCO_(3))`
`Rightarrow M_(NaHCO_(3))=14.4mL`
Thus, 14.4mL of hte 1.25 M `NaHCO_(3)` solution is needed to neutralise 18.0 to neutratlise 18.0mL of the 0.100M HCl solution