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

To find the molarity of the gastric juice, we can use the concept of neutralization reactions. The moles of KOH used in the reaction is equal to the moles of HCl in the gastric juice. By calculating the moles of KOH used and using this information, we can determine the molarity of the gastric juice to be 0.100 M.

Step-by-step explanation:

In order to find the molarity of gastric juice, we can use the concept of neutralization reactions. The balanced chemical equation for the reaction between KOH and HCl is KOH + HCl -> KCl + H2O.
From the equation, we can see that 1 mole of KOH reacts with 1 mole of HCl to form 1 mole of KCl and 1 mole of water. Therefore, the moles of KOH used in the reaction is equal to the moles of HCl in the gastric juice.

First, we need to calculate the moles of KOH used.

moles of KOH = volume of KOH (in L) x molarity of KOH

moles of KOH = (10.0 mL / 1000 mL/L) x 0.300 M = 0.003 moles

Since the moles of KOH is equal to the moles of HCl in the gastric juice, we can use this information to calculate the molarity of the gastric juice.

molarity of HCl = moles of HCl / volume of HCl (in L)

molarity of HCl = 0.003 moles / (30.0 mL / 1000 mL/L) = 0.100 M

Answer 2

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.