Question

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?

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.