Question

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

Answer 1

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:

$C_2H_5NH_{2(aq)$     +     $H^+_{(aq)$      ⇄        $C_2H_5NH_{3(aq)}^+$

Given that concentration of $C_2H_5NH_{2(aq)$ = 10%

i.e 10 g of $C_2H_5NH_{2(aq)$ in 100 ml solution

molar mass = 45.08 g/mol

number of moles = $\frac{10}{45.08}$

= 0.222 mol

Molarity of $C_2H_5NH_{2(aq)$ = 0.222 × $\frac{1000}{100}mL$

= 2.22 M

However, number of moles of $C_2H_5NH_{2(aq)$ in 20 mL can be determined as:

number of moles of $C_2H_5NH_{2(aq)$ = 20 mL × 2.22 M

= $44*10^{-3} mole$

Concentration of $C_2H_5NH_{2(aq)$ = $\frac{44*10^{-3}*1000}{20}$

= 2.22 M

Similarly, The pKa Value of $C_2H_5NH_{2(aq)$  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= $\frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]$

pH = $\frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]$

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.