Answer:
Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6
Bromocresol green, color change from pH = 4.0 to 5.6
Explanation:
The equation for the reaction is :
[tex]C_2H_5NH_2_(_a_q_) + H^+_(_a_q_) --- C_2H_5NH_{3(aq)}^+[/tex]
concentration of [tex]C_2H_5NH_{2(aq)[/tex] = 10%
10 g of [tex]C_2H_5NH_{2(aq)[/tex] in 100 ml solution
molar mass = 45.08 g/mol
number of moles = 10 / 45.08
= 0.222 mol
Molarity of [tex]C_2H_5NH_2(aq) = 0.222 \times \frac{1000}{100}mL[/tex]
= 2.22 M
number of moles of [tex]C_2H_5NH_{2(aq)[/tex] in 20 mL can be determined as:
[tex]= 20 mL \times 2.22 M= 44*10^{-3} mole[/tex]
Concentration of [tex]C_2H_5NH_2(aq) = \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
the pH value at equivalence point is,
[tex]pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C][/tex]
[tex]pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22][/tex]
[tex]pH = 5.21[/tex]
Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6
