calculate the pH at 0ml at the quivalence point and at 40ml in a titration of 25 ml of .120M formic acid with .105M NaOH
(kA of formic acid= 1.8x10^-4)
The key to this type problem is to recognize what is in the solution at the point of calculation.
0 mL must be pure formic acid.
The equivalence point is the salt, sodium formate. so the pH is determined by the hydrolsis (the Kb) of the salt.
At 40.00 mL, (past the equivalence point), it is essentially NaOH.
Does this help get you started? If you have questions, post your work to that point and explain what you don't understand.
To calculate the pH at 0 mL and 40 mL in the titration, we need to determine the concentration of the species involved in the reaction at these points. We'll start by calculating the moles of formic acid (HCOOH) and NaOH in the solution.
1. Moles of formic acid (HCOOH) in 25 mL of 0.120 M solution:
moles of HCOOH = concentration (M) x volume (L)
moles of HCOOH = 0.120 M x 0.025 L
moles of HCOOH = 0.003 mol
2. Moles of NaOH in 40 mL of 0.105 M solution:
moles of NaOH = concentration (M) x volume (L)
moles of NaOH = 0.105 M x 0.040 L
moles of NaOH = 0.0042 mol
Now, we need to determine which reactant is in excess and which is the limiting reactant in the solution.
The balanced chemical equation for the reaction is:
HCOOH + NaOH → HCOONa + H2O
From the equation, we can see that the stoichiometric ratio between HCOOH and NaOH is 1:1. Therefore, whichever reactant is in excess will determine the pH of the solution.
To determine if HCOOH or NaOH is in excess, we need to compare the moles of each reactant. Since the moles of HCOOH (0.003 mol) are less than the moles of NaOH (0.0042 mol), HCOOH is the limiting reactant.
At the equivalence point (the point at which the stoichiometric ratio between the reactants is exactly 1:1), all the limiting reactant (HCOOH) will be consumed.
1. pH at 0 mL (before any reactant is added):
Since we have pure formic acid (HCOOH) with no NaOH added, the pH is determined by the concentration of HCOOH alone. To calculate the pH, we'll use the dissociation constant (Ka) of formic acid (1.8 x 10^-4):
Ka = [H+][HCOO-] / [HCOOH]
Since HCOOH is a weak acid, we can assume that most of it will be undissociated (x) and neglect-x approximation:
Ka = (x)(x) / (0.120 - x)
x^2 = 1.8 x 10^-4 * (0.120 - x)
solve for x (concentration of H+), which will give us the pH at 0 mL.
2. pH at 40 mL (at the equivalence point):
At the equivalence point, all the HCOOH has reacted with NaOH, resulting in the formation of a salt, sodium formate (HCOONa). To determine the pH at this point, we need to calculate the concentration of the resulting salt, HCOONa, using the stoichiometry of the reaction.
Since the stoichiometric ratio between HCOOH and HCOONa is 1:1, we can calculate the concentration of HCOONa by dividing the moles of NaOH (0.0042 mol) by the volume (L) at 40 mL (0.040 L).
Concentration of HCOONa = moles of HCOONa / volume (L)
Concentration of HCOONa = 0.0042 mol / 0.040 L
Compute the concentration to determine the pH using the appropriate dissociation constant (if applicable).
Remember that the pH scale ranges from 0 to 14, with pH values below 7 being acidic, pH 7 being neutral, and pH values above 7 being basic.