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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they all have one thing in common and that its all made up of atoms. When these components are active it creates energy
Answer:
Part 1:sodium
rubidium
Part 2: protons neutrons and electrons are all 12
The number of protons is equal to the no. of neutrons from the electronic arrangement of magnesium and the no. of electrons is got from the atomic no. of magnesium
Answer:1) Volume of 
required is 55.98 mL.
2) 0.62577 grams of 
is produced.
Explanation:
1) Molarity of 
Volume of 
Molarity of 
Volume of 
According to reaction, 1 mole of 
reacts with 3 mole of , then, 0.0041985 moles of will react with:
moles of that is 0.0125955 moles.
Volume of required is 55.98 mL.
2)
Number of moles of 
According to reaction, 3 moles of gives 1 mole of , then 0.004485 moles of will give:
moles of that is 0.001495 moles.
Mass of =
Moles of × Molar Mass of
= 0.001495 moles × 418.58 g/mol = 0.62577 g
0.62577 grams of is produced.
