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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Answer: 0.8541 grams of HCl will be required.
Explanation: Moles can be calculated by using the formula:
Given mass of 
= 0.610 g
Molar mass of = 78 g/mol
Number of moles of = 0.0078 moles
The reaction between and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.
Chemical equation for the above reaction follows:
By Stoichiometry,
1 mole of reacts with 3 moles of HCl
So, 0.0078 moles of will react with 
= 0.0234 moles
Mass of HCl is calculated by using the mole formula, we get
Molar mass of HCl = 36.5 g/mol
Putting values in the equation, we get
Mass of HCl required will be = 0.8541 grams
