Answer :
Answer: The pH of resulting solution is 10.893
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
- For ethylamine:
Molarity of ethylamine solution = 0.25 M
Volume of solution = 80 mL
Putting values in above equation, we get:
[tex]0.25M=\frac{\text{Moles of ethylamine}\times 1000}{80mL}\\\\\text{Moles of ethylamine}=\frac{0.25\times 80}{1000}=0.02mol[/tex]
- For HCl:
Molarity of HCl = 0.100 M
Volume of solution = 20.0 mL
Putting values in above equation, we get:
[tex]0.100M=\frac{\text{Moles of HCl}\times 1000}{20.0mL}\\\\\text{Moles of HCl}=\frac{0.100\times 20}{1000}=0.002mol[/tex]
- For [tex]C_2H_5NH_3Cl[/tex]:
Molarity of [tex]C_2H_5NH_3Cl[/tex] solution = 0.25 M
Volume of solution = 80 mL
Putting values in above equation, we get:
[tex]0.25M=\frac{\text{Moles of }C_2H_5NH_3Cl\times 1000}{80mL}\\\\\text{Moles of }=\frac{0.25\times 80}{1000}=0.02mol[/tex]
The chemical reaction for ethylamine and HCl follows the equation:
[tex]C_2H_5NH_2+HCl\rightarrow C_2H_5NH_3Cl[/tex]
Initial: 0.02 0.002 0.02
Final: 0.018 - 0.022
Volume of solution = 20.0 + 80.0 = 100 mL = 0.100 L (Conversion factor: 1 L = 1000 mL)
To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pOH=pK_b+\log(\frac{[salt]}{[base]})[/tex]
[tex]pOH=pK_b+\log(\frac{[C_2H_5NH_3Cl]}{[C_2H_5NH_2]})[/tex]
We are given:
[tex]pK_b[/tex] = negative logarithm of base dissociation constant of ethylamine = [tex]-\log(9.5\times 10^{-4})=3.02[/tex]
[tex][C_2H_5NH_3Cl]=\frac{0.022}{0.100}[/tex]
[tex][C_2H_5NH_2]=\frac{0.018}{0.100}[/tex]
pOH = ?
Putting values in above equation, we get:
[tex]pOH=3.02+\log(\frac{0.022/0.100}{0.018/0.100})\\\\pOH=3.107[/tex]
To calculate pH of the solution, we use the equation:
[tex]pH+pOH=14\\pH=14-3.107=10.893[/tex]
Hence, the pH of the solution is 10.893
The pH of the solution is 10.9
Data;
- Volume of buffer = 80mL
- Volume of HCL = 20.0mL
- conc. of C2H5NH2 = 0.25M
- conc. of C2H5NH3Cl = 0.25
- Kb of C2H5NH2 = 9.5*10^-4
pH of a Solution
The pH of buffer can be calculated by using Henderson-Hasselbalch's equation
[tex]pOH = _pKb+ log \frac{[salt]}{[base]}[/tex]
The initial moles of salt present is calculated as
[tex]0.25 * 80*10^-^3 = 0.02mmoles[/tex]
The initial moles of base present is calculated as
[tex]0.25*80*10^-^3 = 20mmoles[/tex]
On adding HCl the following reaction will occurs
[tex]C_2H_5NH_2 + HCl \to C_2H_5NH_3Cl[/tex]
This will lead to formation of extra moles of salt that is equal to moles of acid added and eventually lead to decrease in number of moles of base by equal measure.
Moles of HCl added is
[tex]moles of HCL= 0.1 * 20 * 10^-^3 = 2mmoles[/tex]
Adding the value
Moles of salt present = 20 + 2 = 22mmoles
Subtracting the value
Moles of base left = 20-2 = 18mmoles
Now using Henderson-Hasselbalch's equation we can calculate the pOH of solution
[tex]pKb = -logKb = -log (9.5*10^-^4) = 3.02[/tex]
The pOH of the base can be calculated as
[tex]pOH = 3.02 + log (\frac{22}{18}) = 3.107[/tex]
Using the above, we can solve for the pH of the solution.
[tex]pH = 14 - pOH = 10.893[/tex]
The pH of the solution is 10.9
Learn more on pH of a solution using Henderson-Hasselbalch equation here;
https://brainly.com/question/13557815