A 100.0 mL of 0.200 M methylamine, CH3NH2, is titrated with 0.100 M HCl. Calculate the pH after 70.0 mL of 0.100 M HCl has been added.
The trick to these problems is to determine what is present in the solution at the point of your calculation.
CH3NH2 + HCl ==> CH3NH2*HCl.
moles CH3NH2 = M x L = 0.200 x 0.1 = 0.02
moles HCl = M x L = 0.100 x 0.070 = 007.
So you will have an excess of CH3NH2 (all of it is not neutralized) and all of the HCl has been used. You have 0.007 moles of the salt formed. What kind of a solution is this? It is a buffered solution. You have a weak base and its salt present. Use the Henderson-Hasselbalch equation.
pH = pKa + log (base/acid) = ??
To calculate the pH after adding 70.0 mL of 0.100 M HCl to the solution containing 100.0 mL of 0.200 M methylamine, we need to consider the reaction between the weak base, methylamine (CH3NH2), and the strong acid, hydrochloric acid (HCl). This reaction will form the conjugate acid of the weak base, methylammonium (CH3NH3+), and the chloride ion (Cl-).
First, let's determine the limiting reagent in this reaction.
Since the initial volume of methylamine is 100.0 mL, and the volume of HCl added is 70.0 mL, the limiting reagent is the one that will be completely consumed first.
The balanced chemical equation for the reaction is as follows:
CH3NH2 + HCl -> CH3NH3+ + Cl-
The stoichiometry of the reaction tells us that one mole of methylamine reacts with one mole of HCl, so the limiting reagent is the one in smaller molar amount. To calculate the limiting reagent, we can use the formula:
n = C x V
Where "n" is the number of moles, "C" is the concentration, and "V" is the volume.
The number of moles of methylamine (CH3NH2) can be calculated as:
n(CH3NH2) = 0.200 M x 0.100 L = 0.020 moles
The number of moles of HCl can be calculated as:
n(HCl) = 0.100 M x 0.070 L = 0.007 moles
Since the number of moles of HCl (0.007 moles) is lower than the number of moles of methylamine (0.020 moles), HCl is the limiting reagent.
Now, let's calculate the number of moles of HCl remaining after the reaction is complete:
n(HCl remaining) = n(HCl initially) - n(HCl used)
n(HCl used) can be calculated as:
n(HCl used) = 0.100 M x 0.070 L = 0.007 moles
n(HCl remaining) = 0.007 moles - 0.007 moles = 0 moles
Since all the HCl is used up in the reaction, we can consider the remaining volume after the reaction to be the sum of the initial volumes of methylamine and HCl:
Volume remaining = 100.0 mL + 70.0 mL = 170.0 mL
Now, let's calculate the concentration of CH3NH3+ in the solution. Since the volume increased to 170.0 mL, we can calculate the new concentration as:
C' = n / V
C' (CH3NH3+) = 0.007 moles / 0.170 L = 0.041 M
To determine the pH, we need to consider the ionization of CH3NH3+ in water. Methylammonium is the conjugate acid of the weak base methylamine, and it will partially dissociate in water, forming hydronium ions (H3O+) and the conjugate base (CH3NH2).
The equilibrium expression for the ionization of methylammonium (CH3NH3+) can be written as:
CH3NH3+ + H2O ⇌ CH3NH2 + H3O+
The Ka expression is:
Ka = [CH3NH2] [H3O+] / [CH3NH3+]
Since the initial concentration of methylamine is 0.200 M and the concentration of hydronium ions is not given, we will assume that it is negligible initially.
Now we can calculate the concentration of hydronium ions using the equilibrium expression:
Ka = [CH3NH2] [H3O+] / [CH3NH3+]
0.041 = (0.200) (x) / (0.200)
x = 0.041
Therefore, the concentration of hydronium ions (H3O+) is 0.041 M.
To calculate the pH, we will use the formula:
pH = -log10[H3O+]
pH = -log10(0.041)
pH ≈ 1.39
So, the pH after adding 70.0 mL of 0.100 M HCl to the solution containing 100.0 mL of 0.200 M methylamine is approximately 1.39.
To calculate the pH after 70.0 mL of 0.100 M HCl has been added to the 100.0 mL of 0.200 M methylamine solution, we need to consider the reaction between the methylamine (CH3NH2) and HCl.
First, let's write down the balanced chemical equation for the reaction between methylamine and HCl:
CH3NH2 + HCl → CH3NH3+ + Cl-
The methylamine (CH3NH2) acts as a weak base, and HCl is a strong acid. The reaction produces the methylammonium ion (CH3NH3+) and chloride ion (Cl-).
Since methylamine acts as a weak base, we can consider it as a weak acid instead (CH3NH3+). The dissociation of methylamine can be represented by the following equation:
CH3NH3+ + H2O ⇌ CH3NH2 + H3O+
The Ka value for methylamine is 2.3 x 10^-11.
Now, let's calculate the moles of methylamine initially present in the solution before the titration:
Moles of CH3NH2 = Volume (L) x Concentration (M)
= 0.100 L x 0.200 M
= 0.020 mol
Next, let's calculate the moles of HCl added during the titration:
Moles of HCl = Volume (L) x Concentration (M)
= 0.070 L x 0.100 M
= 0.007 mol
By the stoichiometry of the reaction, we can see that 1 mole of HCl reacts with 1 mole of CH3NH2 (methylamine).
Since the moles of HCl added (0.007 mol) are less than the moles of methylamine initially present (0.020 mol), we know that excess methylamine remains in solution. Therefore, we can treat it as a buffer solution.
To calculate the concentrations of CH3NH3+ (methylammonium ion) and CH3NH2 (methylamine) in the buffer solution, we use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH = The pH of the solution
pKa = The -logKa value of the weak acid (CH3NH3+)
[A-] = Concentration of the conjugate base (CH3NH2)
[HA] = Concentration of the weak acid (CH3NH3+)
Since the concentration of CH3NH2 (methylamine) is known (0.020 mol/0.170 L), we need to determine the concentration of CH3NH3+ (methylammonium ion).
To do this, we subtract the moles of the weak acid (CH3NH3+) formed by the reaction between HCl and methylamine from the initial moles of methylamine.
Moles of CH3NH3+ = moles of methylamine initially - moles of HCl added
= 0.020 mol - 0.007 mol
= 0.013 mol
Concentration of CH3NH3+ = Moles / volume
= 0.013 mol / 0.170 L
= 0.076 M
Now, we can plug the values into the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([A-]/[HA])
= -log(2.3 x 10^-11) + log(0.076/0.020)
≈ -log(2.3 x 10^-11) + log(3.8)
≈ 10.64 + 0.58
≈ 11.22
Therefore, the pH after 70.0 mL of 0.100 M HCl has been added is approximately 11.22.