Use the Henderson-Hasselbalch equation to calculate the pH of each solution:
a solution that contains 1.05% C2H5NH2 by mass and 1.10% C2H5NH2Br by mass
Percent by mass means grams solute/100 g solution. Since each solution is approximately the same, I would ignore the small difference and convert 1.05 g and 1.10 g to moles and substitute moles into the HH equation. Show your work if you get stuck.
How many grams in 1.26 x 10 -4 mol of HC2H3O2
To calculate the pH of the solution containing C2H5NH2 and C2H5NH2Br, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Given:
Mass of C2H5NH2 = 1.05%
Mass of C2H5NH2Br = 1.10%
First, we need to convert the mass percentages into concentrations (molarities). The molar mass of C2H5NH2 (ethylamine) is 45.08 g/mol, and the molar mass of C2H5NH2Br (ethylammonium bromide) is 137.02 g/mol.
To convert the mass percentages into concentrations, we can assume 100 g of the solution, so we have:
Mass of C2H5NH2 = (1.05/100) * 100 g = 1.05 g
Mass of C2H5NH2Br = (1.10/100) * 100 g = 1.10 g
Now, we can calculate the moles of C2H5NH2 and C2H5NH2Br using their respective molar masses:
Moles of C2H5NH2 = 1.05 g / 45.08 g/mol = 0.0233 mol
Moles of C2H5NH2Br = 1.10 g / 137.02 g/mol = 0.00803 mol
Next, we can calculate the concentrations of C2H5NH2 and C2H5NH2Br by dividing the moles by the total volume of the solution. Since the volume is not given, we'll assume it to be 1 L for simplicity:
Concentration of C2H5NH2 = 0.0233 mol / 1 L = 0.0233 M
Concentration of C2H5NH2Br = 0.00803 mol / 1 L = 0.00803 M
Now, let's denote C2H5NH2 as the base (A-) and C2H5NH2Br as the acid (HA). The pKa for the ethylamine-ethylammonium bromide system is approximately 10.73 (varies slightly depending on temperature and ionic strength).
Substituting the values into the Henderson-Hasselbalch equation:
pH = 10.73 + log([0.00803]/[0.0233])
Now, calculate the ratio: [0.00803]/[0.0233] = 0.345.
pH = 10.73 + log(0.345)
Using a calculator, we find that log(0.345) ≈ -0.462.
pH = 10.73 - 0.462
Finally, we can calculate the pH:
pH ≈ 10.268
Therefore, the pH of the solution containing 1.05% C2H5NH2 by mass and 1.10% C2H5NH2Br by mass is approximately 10.268.
To calculate the pH of a solution using the Henderson-Hasselbalch equation, we need to know the concentrations of the acid and its conjugate base. In this case, we have a solution of C2H5NH2 (ethylamine) and C2H5NH2Br (ethylamine hydrobromide).
The first step is to calculate the concentrations of C2H5NH2 and C2H5NH2Br in the solution. To do this, we need to know their molecular weights and assume 100g of solution.
The molecular weight of C2H5NH2 is:
C: 12.01g/mol + 2 * H: 1.01g/mol + 5 * H: 1.01g/mol = 17.04 g/mol
The molecular weight of C2H5NH2Br is:
C: 12.01g/mol + 2 * H: 1.01g/mol + 5 * H: 1.01g/mol + Br: 79.90g/mol = 94.93 g/mol
Now we can calculate the mass of C2H5NH2 in the solution:
1.05% * 100g = 1.05g of C2H5NH2
And the mass of C2H5NH2Br in the solution:
1.10% * 100g = 1.10g of C2H5NH2Br
Next, we can calculate the number of moles of C2H5NH2:
moles of C2H5NH2 = mass of C2H5NH2 / molecular weight of C2H5NH2
moles of C2H5NH2 = 1.05g / 17.04 g/mol
Similarly, we can calculate the number of moles of C2H5NH2Br:
moles of C2H5NH2Br = mass of C2H5NH2Br / molecular weight of C2H5NH2Br
moles of C2H5NH2Br = 1.10g / 94.93 g/mol
Now we have the concentrations of C2H5NH2 and C2H5NH2Br in the solution, which we can use in the Henderson-Hasselbalch equation:
pH = pKa + log([base] / [acid])
In this equation, [base] is the concentration of the conjugate base (C2H5NH2) and [acid] is the concentration of the acid (C2H5NH2Br). The pKa is the logarithmic value of the acid dissociation constant for the C2H5NH2/C2H5NH2Br acid-base system.
The pKa value for C2H5NH2/C2H5NH2Br is not provided, so we cannot directly calculate the pH without that information. The pKa is necessary to solve the Henderson-Hasselbalch equation and determine the pH.
In summary, to calculate the pH of the solution containing 1.05% C2H5NH2 and 1.10% C2H5NH2Br, we need to know the pKa value for the C2H5NH2/C2H5NH2Br acid-base system.