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A 25.00 ml sample of gastric juice is titrated with a 0.0201 m NaOH solution. The titration to the equivalence point requires 26.7 ml NaOH solution. if the equation for the reaction is: hcl(aq) + NaOH (aq) = NaCl (aq) + H2O (l)

Accepted Answer

The concentration of HCl in the gastric juice can be calculated as follows:
1. Moles of NaOH:
- Moles of NaOH = Molarity x Volume (in liters)
- Moles of NaOH = 0.0201 M x 0.0267 L = 0.000538 mol
2. Moles of HCl:
- From the balanced chemical equation, 1 mole of HCl reacts with 1 mole of NaOH.
- Therefore, moles of HCl = moles of NaOH = 0.000538 mol
3. Concentration of HCl:
- Concentration of HCl = Moles of HCl / Volume of gastric juice (in liters)
- Concentration of HCl = 0.000538 mol / 0.0250 L = 0.0215 M
Therefore, the concentration of HCl in the gastric juice is 0.0215 M.

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