Answer:
[tex]\boxed{\text{36 mL}}[/tex]
Explanation:
1. Write the balanced chemical equation.
[tex]\rm HCOOH + NaOH $ \longrightarrow$ HCOONa + H$_{2}$O[/tex]
2. Calculate the moles of HCOOH
[tex]\text{Moles of HCOOH} =\text{20.00 mL HCOOH } \times \dfrac{\text{0.20 mmol HCOOHl}}{\text{1 mL HCOOH}} = \text{4.00 mmol HCOOH}[/tex]
3. Calculate the moles of NaOH.
[tex]\text{Moles of NaOH = 4.00 mmol HCOOH } \times \dfrac{\text{1 mmol NaOH} }{\text{1 mmol HCOOH}} = \text{4.00 mmol NaOH}[/tex]
4. Calculate the volume of NaOH
[tex]c = \text{4.00 mmol NaOH } \times \dfrac{\text{1 mL NaOH }}{\text{0.1105 mmol NaOH }} = \textbf{36 mL NaOH }\\\\\text{The titration will require }\boxed{\textbf{36 mL of NaOH}}[/tex]
The volume of NaOH required to reach the end point is 36.2 mL
We'll begin by writing the balanced equation for the reaction.
HCOOH + NaOH —> HCOONa + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, NaOH (nB) = 1
From the question given above,
Molarity of base, NaOH (Mb) = 0.1105 M
Volume of acid, HCOOH (Va) = 20 mL
Molarity of acid, HCOOH (Ma) = 0.2 M
MaVa / MbVb = nA/nB
(0.2 × 20) / (0.1105 × Vb) = 1
4 / (0.1105 × Vb) = 1
Cross multiply
0.1105 × Vb = 4
Divide both side by 0.1105
Vb = 4 / 0.1105
Thus, the volume of the base, NaOH is 36.2 mL
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