You are given a solution of HCOOH (formic acid) with an approximate concentration of 0.20 M and you will titrate with a 0.1105 M NaOH. If you add 20.00 mL of HCCOOH to the beaker before titrating, approximately what volume of NaOH will be required to reach the end point?
36.2 mL. I don't know how to get the answer but I know that that's what it is.
I assume that's a typo on the second HCOOH.
mols HCOOH = M x L = ?
mols NaOH = the same
M NaOH = mols NaOH/L NaOH. YOu know mols and M, solve for L
To determine the volume of NaOH required to reach the end point, we can use the following balanced chemical equation for the reaction between formic acid (HCOOH) and sodium hydroxide (NaOH):
HCOOH + NaOH → HCOONa + H2O
The stoichiometry of the reaction implies that 1 mole of HCOOH reacts with 1 mole of NaOH to produce 1 mole of HCOONa.
Given that the concentration of the HCOOH solution is 0.20 M and the volume added is 20.00 mL (which we'll convert to liters), we can calculate the number of moles of HCOOH added:
moles of HCOOH = concentration × volume
= 0.20 M × 0.02000 L
= 0.004 moles
Since the balanced equation shows a 1:1 ratio between HCOOH and NaOH, 0.004 moles of NaOH will be required to reach the endpoint.
To find the volume of NaOH required, we can use its concentration of 0.1105 M:
volume of NaOH = moles of NaOH / concentration
= 0.004 moles / 0.1105 M
≈ 0.036 L
Converting the volume from liters to milliliters, we find that approximately 36.0 mL of NaOH will be required to reach the end point.
To determine the approximate volume of NaOH required to reach the endpoint, we can use the following equation:
Molarity (HCOOH) × Volume (HCOOH) × Number of H+ ions (HCOOH) = Molarity (NaOH) × Volume (NaOH) × Number of OH- ions (NaOH)
First, let's calculate the number of moles of formic acid (HCOOH) in the solution we're given:
Molarity (HCOOH) = 0.20 M
Volume (HCOOH) = 20.00 mL = 0.02000 L (converted to liters)
Number of moles (HCOOH) = Molarity × Volume = 0.20 M × 0.02000 L = 0.0040 mol (moles)
Next, since formic acid (HCOOH) is a monoprotic acid, it donates one hydrogen ion (H+) in a reaction. Therefore, the number of H+ ions is 1.
Now, we know the molarity of NaOH is 0.1105 M. Let's assume the volume of NaOH required to reach the endpoint is V (in liters).
Number of OH- ions (NaOH) = 1 (since NaOH is a strong base)
Plugging all this information into the equation:
0.0040 mol (HCOOH) × 1 (H+) = 0.1105 M × V (NaOH) × 1 (OH-)
0.0040 = 0.1105 V
Solving for V:
V ≈ 0.0040 / 0.1105 ≈ 0.0362 L
Therefore, approximately 0.0362 liters of NaOH (36.2 mL) will be required to reach the endpoint.