The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
Learn more about titration:
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An ionized has consisting of positive ions and free elections in proportion resulting in more or less no overall electric charge
Answer:
1.242 g/mL
Explanation:
Step 1: Given data
Mass of the empty container (m₁): 80.21 g
Mass of the filled container (m₂): 105.22 g
Volume of the unknown liquid (V): 20.14 mL
Step 2: Calculate the mass of the liquid
The mass of the liquid is equal to the difference between the mass of the filled container and the mass of the empty container.

Step 3: Calculate the density of the unknown liquid
The density of the liquid is equal to its mass divided by its volume.

Answer:

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