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A 20.0 mL sample of gastric juice was taken from a patient suspected of having hypochloridia (low stomach acid). The sample was taken several hours after a meal so there was no ingested food or drink, and it was assumed no buffers were present. The sample required 6.3 mL of 0.0011 M NaOH to titrate the sample to neutrality.

What was the pH of the gastric juice?

  • biochemistry -

    mols NaOH = M x L = ?
    mols HCl(in the gastric juice) = mols NaOH
    M HCl = mols HCl/L HCl. L = 0.020.
    Then convert M to pH.
    pH = -log(H^+).

  • biochemistry -


  • biochemistry -


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