20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator would be most suitable for this titration? The pKa of CH3CH2NH3+ is 10.75.
Answer:
Bromocresol green, color change from pH = 4.0 to 5.6
Explanation:
The equation for the reaction is as follows:
[tex]C_2H_5NH_{2(aq)[/tex] + [tex]H^+_{(aq)[/tex] ⇄ [tex]C_2H_5NH_{3(aq)}^+[/tex]
Given that concentration of [tex]C_2H_5NH_{2(aq)[/tex] = 10%
i.e 10 g of [tex]C_2H_5NH_{2(aq)[/tex] in 100 ml solution
molar mass = 45.08 g/mol
number of moles = [tex]\frac{10}{45.08}[/tex]
= 0.222 mol
Molarity of [tex]C_2H_5NH_{2(aq)[/tex] = 0.222 × [tex]\frac{1000}{100}mL[/tex]
= 2.22 M
However, number of moles of [tex]C_2H_5NH_{2(aq)[/tex] in 20 mL can be determined as:
number of moles of [tex]C_2H_5NH_{2(aq)[/tex] = 20 mL × 2.22 M
= [tex]44*10^{-3} mole[/tex]
Concentration of [tex]C_2H_5NH_{2(aq)[/tex] = [tex]\frac{44*10^{-3}*1000}{20}[/tex]
= 2.22 M
Similarly, The pKa Value of [tex]C_2H_5NH_{2(aq)[/tex] is given as 10.75
pKb value will be: 14 - pKa
= 14 - 10.75
= 3.25
Finally, the pH value at equivalence point is:
pH= [tex]\frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C][/tex]
pH = [tex]\frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22][/tex]
pH = 5.21
∴
The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6.
