Respuesta :
The question is incomplete, here is the complete question:
Assuming that all the [tex]H^+[/tex] comes from HCl, how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid? Volume = 500 mL pH= 2
Answer: The mass of sodium hydrogen carbonate needed to completely neutralize stomach acid is 0.42 grams
Explanation:
To calculate the hydrogen ion concentration of the solution, we use the equation:
[tex]pH=-\log[H^+][/tex]
We are given:
pH = 2
Putting values in above equation, we get:
[tex]2=-\log[H^+][/tex]
[tex][H^+]=10^{-2}M[/tex]
- To calculate the number of moles for given molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Molarity of hydrogen ions = 0.01 M
Volume of solution = 500 mL = 0.5 L (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
[tex]0.01M=\frac{\text{Moles of hydrogen ions}}{0.5L}\\\\\text{Moles of hydrogen ions}=(0.01mol/L\times 0.5L)=0.005mol[/tex]
The chemical equation for the reaction of HCl and sodium hydrogen carbonate follows:
[tex]HCl+NaHCO_3\rightarrow NaCl+H_2CO_3[/tex]
By Stoichiometry of the reaction:
1 mole of HCl reacts with 1 mole of sodium hydrogen carbonate
So, 0.005 moles of HCl will react with = [tex]\frac{1}{1}\times 0.005=0.005mol[/tex] of sodium hydrogen carbonate
- To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of sodium hydrogen carbonate = 0.005 moles
Molar mass of sodium hydrogen carbonate = 84 g/mol
Putting values in above equation, we get:
[tex]0.005mol=\frac{\text{Mass of sodium hydrogen carbonate}}{84g/mol}\\\\\text{Mass of sodium hydrogen carbonate}=(0.005mol\times 84g/mol)=0.42g[/tex]
Hence, the mass of sodium hydrogen carbonate needed to completely neutralize stomach acid is 0.42 grams
