Number of moles of NaOH = V(NaOH) * M(NaOH)= 0.150 L * 0.1 moles/L = 0.015 moles
Number of moles of formic acid, HCOOH = V(HCOOH) * M(HCOOH) = 0.200 L * 0.1 moles/L = 0.020 moles
Here, the limiting reagent is NaOH
The reaction is represented as:
HCOOH + NaOH ↔HCOONa + H2O
Moles of HCOONa formed = Moles of the limiting reagent, NaOH = 0.015 moles
Moles of HCOOH remaining = 0.020-0.015 = 0.005 moles
Total final volume is given as 1 L
Therefore: [HCOOH] = 0.005 moles/1 L = 0.005 M
[HCOONa] = 0.015/1 = 0.015 M
pKa of HCOOH = 3.74
As per Henderson-Hasselbalch equation
pH = pka + log[HCOONa]/[HCOOH] = 3.74+log[0.015/0.005] = 4.22
Therefore, pH of the final solution = 4.22
Formic acid is the simplest carboxylic acid with a structure of HCOOH and has a pka of 3.75. The pka refers to the acidity of the molecule, which in this example refers to the molecules ability to give up the proton of the O-H. A decrease in the pka value corresponds to an increase in acidity, or an increase in the ability to give up a proton. When an acid gives up a proton, the remaining anionic species (in this case HCOO-) is called the conjugate base, and an increase in the stability of the conjugate base corresponds to an increase in acidity.
The pka of a carboxylic can be affected greatly by the presence of various functional groups within its structure. An example of an inductive effect changing the pka can be shown with trichloroacetic acid, Cl3CCOOH. This molecule has a pka of 0.7. The decrease in pka relative to formic acid is due to the presence of the Cl3C- group, and more specifically the presence of the chlorine atoms. The electronegative chlorine atoms are able to withdraw the electron density away from the oxygen atoms and towards themselves, thus helping to stabilize the negative charge and stabilize the conjugate base. This results in an increase in acidity and decrease in pka.
The same Cl3CCOOH example can be used to explain how dipoles can effect the acidity of carboxylic acids. Compared to standard acetic acid, H3CCOOH with a pka of 4.76, trichloroacetic acid is much more acidic. The difference between these structures is the presence of C-Cl bonds in place of C-H bonds. A C-Cl bond is much more polar than a C-H bond, due the large electronegativity of the chlorine atom. This results in a carbon with a partial positive charge and a chlorine with a partial negative charge. In the conjugate base of the acid, where the molecule has a negative charge localized on the oxygen atoms, the dipole moment of the C-Cl bond is oriented such that the partial positive charge is on the carbon that is adjacent to the oxygen atoms containing the negative charge. Therefore, the electrostatic attraction between the positive end of the C-Cl dipole and the negative charge of the anionic oxygen helps to stabilize the entire species. This level of stabilization is not present in acetic acid where there are C-H bonds instead of C-Cl bonds since the C-H bonds do not have a large dipole moment.
To understand how resonance can affect the pka of a species, we can simply compare the pka of a simple alcohol such as methanol, CH3OH, and formic acid, HCOOH. The pka of methanol is 16, suggesting that is is a very weak acid. Once methanol gives up that proton to become the conjugate base CH3O-, the charge cannot be stabilized in any way and is simply localized on the oxygen atom. However, with a carboxylic acid, the conjugate base, HCOO-, can stabilize the negative charge. The lone pair electrons containing the charge on the oxygen atom are able to migrate to the other oxygen atom of the carboxylic acid. The negative charge can now be shared between the two electronegative oxygen atoms, thus stabilizing the charge and decreasing the pka.
You can calculate the excess reactant by subtracting the mass of excess reagent consumed from the total mass of reagent given therefore,
The answer: Theoretical yield is 121.60 g of NH₃
Excess reactant is H₂
Rate limiting reactant is N₂
explanation: 100 g of Nitrogen
100 g of hydrogen
We are required to identify the theoretical yield of the reaction, the excess reactant and the rate limiting reagent.
We first write the equation for the reaction between nitrogen and hydrogen;
N₂ + 3H₂ → 2NH₃
From the reaction 1 mole of nitrogen reacts with 3 moles of Hydrogen gas.
Secondly we determine the moles of nitrogen gas given and hydrogen gas given;
Moles of Nitrogen gas
Moles = Mass ÷ Molar mass
Molar mass of nitrogen gas = 28.0 g/mol
Moles of Nitrogen gas = 100 g ÷ 28 g/mol 3.57 moles
Moles of Hydrogen gas
Molar mass of Hydrogen gas = 2.02 g/mol
Moles = 100 g ÷ 2.02 g/mol
= 49.50 moles
From the mole ratio given by the equation, 1 mole of nitrogen requires 3 moles of Hydrogen gas.
Thus, 3.57 moles of Nitrogen gas requires (3.57 × 3) 10.71 moles of Hydrogen gas.
This means, Nitrogen gas is the rate limiting reagent and hydrogen gas is the excess reactant.
Third calculate the theoretical yield of the reaction.
1 mole of nitrogen reacts to from 2 moles of ammonia gas
Therefore;
Moles of ammonia gas produced = Moles of nitrogen × 2
= 3.57 moles × 2
= 7.14 moles
But; molar mass of Ammonia gas is = 17.03 g/mol
Therefore;
Mass of ammonia gas produced = 7.14 moles × 17.03 g/mol
= 121.59 g
= 121.60 g
Thus, the theoretical amount of ammonia gas produced is 121.60 g
