Question

(a) 6.3 (b) 7.0 (C) 8.1 (d) 5.8 (37. Consider an experiment where 35.0 mL of 0.175 HAc is titrated with 0.25 M NaOH. What is the pH at the equivalence point of this titration? The Ka for HAc is 1.8x10. (a) 5.12 (b) 2.87 (c) 11.13 (d) 8.88 38. 30.0 mL of 0.20 M NH3 is titrated with 30.0 mL of 0.10 M HCI. What is the pH at this point? (Kb = 1.8x10) (b) 4.74 (c) 8.95 (d) 9.56 (a) 9.26

37 and 38

Answer 1

Q37. Correct answer is : (d) 8.88

Q38. Correct answer is : (a) 9.26

Explanation

Q37. concentration HAc = 0.175 M

volume HAc = 35.0 mL

moles HAc = (concentration HAc) * (volume HAc)

moles HAc = (0.175 M) * (35.0 mL)

moles HAc = 6.125 mmol

moles NaOH required = moles HAc

moles NaOH required = 6.125 mmol

volume HAc required = (moles NaOH required) / (concentration NaOH)

volume HAc required = (6.125 mmol) / (0.25 M)

volume HAc required = 24.5 mL

Total volume at equivalence point = (volume HAc) + (volume NaOH)

Total volume at equivalence point = (35.0 mL) + (24.5 mL)

Total volume at equivalence point = 59.5 mL

concentration of Ac^- at equivalence point = (moles HAc) / (Total volume at equivalence point)

concentration of Ac^- at equivalence point = (6.125 mmol) / (59.5 mL)

concentration of Ac^- at equivalence point = 0.01029 M

Ac^- K_b = (K_w) / (HAc K_a)

Ac^- K_b = (1.0 x 10^-14) / (1.8 x 10^-5)

Ac^- K_b = 5.6 x 10^-10

ICE table	Ac-	H2O	$\rightleftharpoons$	HAc	OH-
Initial conc.	0.01029 M	-		0	0
Change	-x	-		+x	+x
Equilibrium	0.01029 M - x	-		+x	+x

K_b = [HAc]_eq[OH^-]_eq/[Ac^-]_eq

5.6 x 10^-10 = [(x) * (x)] / (0.01029 M - x)

Solving for x, x = 2.4 x 10^-6 M

[OH^-] = x = 2.4 x 10^-6 M

pOH = -log[OH^-]

pOH = -log(2.4 x 10^-6 M)

pOH = 5.62

pH = 14 - pOH

pH = 14 - 5.62

pH = 8.38