Question

What is the pH of a solution prepared by adding 120.0 mL of 0.60 M phenylamine (aniline) to 120.0 mL of 0.60 M hydrobromic acid?

	1.10
	4.79
	2.58
	9.21
	11.42

Answer 1

Correct answer is : 2.58

Explanation

Given : concentration of aniline = 0.60 M

volume of aniline = 120.0 mL

moles of aniline = (concentration of aniline) * (volume of aniline)

moles of aniline = (0.60 M) * (120.0 mL)

moles of aniline = 72 mmol

Similarly, moles of HBr = 72 mmol

Since moles of aniline equals moles of HBr, equivalence point is reached and all aniline is converted to anilinium ions

moles of anilinium = moles of aniline = 72 mmol

total volume = (volume of aniline) + (volume of HBr)

total volume = (120.0 mL) + (120.0 mL)

total volume = 240.0 mL

concentration of anilinium = (moles of anilinium) / (total volume)

concentration of anilinium = (72 mmol) / (240.0 mL)

concentration of anilinium = 0.30 M

aniline Kb = 7.4 x 10^-10

anilinium Ka = (Kw) / (Kb)

anilinium Ka = (1 x 10^-14) / (7.4 x 10^-10)

anilinium Ka = 1.35 x 10^-5

ICE table	C6H5NH3+ (aq)		H+ (aq)	C6H5NH2 (aq)
Initial conc.	0.30 M		0	0
Change	-x		+x	+x
Equilibrium conc.	0.30 M - x		+x	+x

Ka = [H+]eq[C6H5NH2]eq / [C6H5NH3+]eq

1.35 x 10^-5 = [(x) * (x)] / (0.30 M - x)

Solving for x, x = 2.0 x 10^-3 M

[H+] = x = 2.0 x 10^-3 M

pH = -log[H+]

pH = -log(2.0 x 10^-3 M)

pH = 2.58