Calculate the pH at the equivalence point for the titration of a solution containing 1250.0 mg of hilariamine (MW = 92.5 g/mol, Kb = 8.1×10−4) with 0.1000 M HCl solution. The volume of the solution at the equivalence point is 175.0 mL.
(I got the answer to be 12.0208 but it seems wrong because pH isn't suppose to be that high)
What about 6.01?
This is an amine so I'm calling it RNH2. At the equivalence point with HCl the salt will be RNH3Cl and the pH will be determined by the hydrolysis of the salt.
RNH3^+ + H2O ==> RNH2 + H3O^+
Ka = (Kw/Kb) = (RNH2)(H3O^+)/(RNH2).
Kw/Kb = 1E-14/8.1E-4 = 1.23E-11
1.23E-11 = (x)(x)/(salt)
(salt) = M = moles/L and moles = g/molar mass
moles = 1.25/92.5 and volume = 0.175; therefore, 1.25/92.5/0.175 = about 0.077
1.23E-11 = (x)(x)/(0.077)
Solve for (H3O^+) and convert to pH. You need to work through it without estimating here and there as I did.
i worked through it just don't know it i did it right...
And your answer should be 6.01
i checked through my work and I found out i didn't do square root for the x^2....thats why my answer was double 6.011.
To calculate the pH at the equivalence point of the titration, we need to determine the concentration of the resulting salt and the hydrolyzed ions in the solution.
The reaction between hilariamine and HCl can be represented as follows:
Hilariamine + HCl -> Hilariammonium Chloride
At the equivalence point, all of the hilariamine is converted into its salt form, which is hilariammonium chloride. In this case, the salt will completely hydrolyze in water to produce hydroxide ions (OH-) and the conjugate acid of hilariammonium, which is ammonium (NH4+).
In order to calculate the concentration of hydroxide ions at the equivalence point, we will use the concept of the Kb value (base dissociation constant) of hilariamine.
The Kb of hilariamine is given as 8.1 × 10^-4. The Kb value allows us to determine the concentration of hydroxide ions produced when the hilariamine salt hydrolyzes.
First, calculate the moles of hilariamine:
1250.0 mg hilariamine = 1.2500 g
Moles of hilariamine = mass / molar mass
Moles of hilariamine = 1.2500 g / 92.5 g/mol
Moles of hilariamine = 0.0135 mol
Since the molar concentration of HCl is given as 0.1000 M, we can assume that 0.1000 moles of HCl reacts with 0.1000 moles of hilariamine to reach the equivalence point.
Now, calculate the volume of HCl reacted:
0.1000 moles HCl = 0.1000 M x Volume (L)
Volume (L) = 0.1000 moles HCl / 0.1000 M
Volume (L) = 1 L
Since we are given the volume of the solution at the equivalence point as 175.0 mL, we have to convert it to liters:
Volume (L) = 175.0 mL / 1000 mL/L
Volume (L) = 0.175 L
Now, we can calculate the concentration of the resulting hilariammonium chloride salt:
Concentration of hilariammonium chloride = moles / volume
Concentration of hilariammonium chloride = 0.1000 moles / 0.175 L
Concentration of hilariammonium chloride = 0.5714 M
Since hilariammonium chloride completely ionizes in water, the concentration of hydroxide ions and ammonium ions will be the same as the concentration of hilariammonium chloride, which is 0.5714 M.
Now, we can calculate the pOH at the equivalence point:
pOH = -log[OH-]
pOH = -log(0.5714)
pOH = 0.2429
Finally, to get the pH at the equivalence point, we can use the fact that pH + pOH = 14:
pH = 14 - pOH
pH = 14 - 0.2429
pH = 13.7571
Therefore, the pH at the equivalence point for the given titration is approximately 13.7571.