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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Explanation:
Since liquid isopropanol is a polar liquid and water is also a polar solvent. So, when both of them are added together then according to the like dissolves like principle they get dissolved.
At the molecular level, the polar molecules of isopropanol get attracted towards the polar molecules of water at the surface of water.
As a result, water molecules get surrounded by isopropanol. Thus, water molecules enter the solution and evenly spread into the solution.
Hi, thank you for posting your question here at Brainly.
The acid dissociation constant, Ka, is an equilibrium constant that measure the strength of an acid. It is a ratio of the concentration of the products (salt and water) and the reactants (acid). The higher the Ka, the more tendency it is to favor the product side, which means more tendency to donate H+ ions. This is exactly the definition of a strong acid (high H+ ionized).
Thus, the answer is letter D.
