If 10.0 ml of 0.300 m koh are required to neutralize 30.0 ml of gastric juice (hcl), what is the molarity of the gastric juice?

Question

Stanley Reese

Chemistry

14 March, 03:34

If 10.0 ml of 0.300 m koh are required to neutralize 30.0 ml of gastric juice (hcl), what is the molarity of the gastric juice?

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Answers (1)

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Felipe Lin · Answer 1

Answer is: the molarity of the gastric juice is 0.100 M.

Chemical reaction: KOH + HCl → KCl + H₂O.

V (KOH) = 10.0 mL : 1000 ml/L = 0.01 L.

c (KOH) = 0.300 M.

V (HCl) = 30.0 mL = 0.03 L.

From chemical reaction: n (KOH) = n (HCl).

V (KOH) · c (KOH) = V (HCl) · c (HCl).

0.01 L · 0.3 M = 0.03 L · c (HCl).

c (HCl) = 0.003 M·L: 0.03 L.

c (HCl) = 0.1 M.