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The concentration of HCl when released to the stomach cavity is diluted to between 0.01 and...

The concentration of HCl when released to the stomach cavity is diluted to between 0.01 and 0.001 M.

How many grams of NaHCO3 are required to neutralize 14.0 mL of a solution having a pH of 1.8? (Express your answer with the appropriate units.)

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Answer #1

   PH = 1.8

-log[H+] = 1.8

     [H+] = 10-1.8

             = 0.0158M

no of moles of H+   = molarity * volume in L

                               = 0.0158*0.014   = 0.0002212 moles

NaHCO3 + H+ -------> Na+ + CO2 + H2O

no of moles of NaHCO3 = no of moles of H+

no of moles of NaHCO3 = 0.0002212moles

mass of NaHCO3    = no of moles * gram molar mass

                               = 0.0002212*84   = 0.0186g >>>> answer

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