Question

Assuming that all the H comes from HCl, how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid?

Answer 1

The question is incomplete, here is the complete question:

Assuming that all the
`H^+` 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

Step-by-step explanation:

To calculate the hydrogen ion concentration of the solution, we use the equation:

`pH=-\log[H^+]`

We are given:

pH = 2

Putting values in above equation, we get:

`2=-\log[H^+]`

`[H^+]=10^(-2)M`

• To calculate the number of moles for given molarity of solution, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}`

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:

`0.01M=\frac{\text{Moles of hydrogen ions}}{0.5L}\\\\\text{Moles of hydrogen ions}=(0.01mol/L* 0.5L)=0.005mol`

The chemical equation for the reaction of HCl and sodium hydrogen carbonate follows:

`HCl+NaHCO_3\rightarrow NaCl+H_2CO_3`

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 =
`(1)/(1)* 0.005=0.005mol` of sodium hydrogen carbonate

• To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Moles of sodium hydrogen carbonate = 0.005 moles

Molar mass of sodium hydrogen carbonate = 84 g/mol

Putting values in above equation, we get:

`0.005mol=\frac{\text{Mass of sodium hydrogen carbonate}}{84g/mol}\\\\\text{Mass of sodium hydrogen carbonate}=(0.005mol* 84g/mol)=0.42g`

Hence, the mass of sodium hydrogen carbonate needed to completely neutralize stomach acid is 0.42 grams