Question

In a hospital laboratory, a 10.0 mL sample of gastric juice, obtained several hours after a meal, was titrated with 0.1 M NaOH to neutrality; 7.2 mL of NaOH was required. The patient’s stomach contained no ingested food or drink. Hence assume that no buffers were present. What was the pH of the gastric juice?

Answer 1

Explanation:

${ \huge{ \texttt{ \fcolorbox{lightblue}{black}{Aɳʂɯҽr}}}}$

HCl + NaOH ------->NaCl + H2O Volume of gastric juice (HCl) = 10.0 mL = 0.01 L letthe molarity of HCl

_____________________

@MɪssSᴘᴀʀᴋʟɪɴɢ❤️