TBL-iRAT CHEM 1308 Spr 2016 Ch.17: Acid/Base, Solubility equilibria (Burdge\'s b

Question

TBL-iRAT CHEM 1308 Spr 2016 Ch.17: Acid/Base, Solubility equilibria (Burdge's book) Group No. This open quiz/guidance/iRAT must be turned in before the in-class tRAT activity Name 1. The main ingredient of vinegar is acetic acid (HAe) that will dissociate into acetate (Ae') and hydrogen ion (H) in water. KHAcroom semp)1.8x10 (a), acetic acid into pure water to make its initial concentration [HAch 0.001 M. Then, what is its(H.h ? What is the reaction extent (the fraction of the reacted reactant) for the HAc dissociation? This reaction extent is in addition to K. This refers to the ICE Table application. (b). What is the conjugate base of HAc? What is the hydrolysis constant of that conjugate base (It should not be called as the base's "dissociation constant")? Adding 0.40 g NaOH 100 ml of 20 M HAc resulted in a complete dissolution. What is the pH of the solution at its equilibrium? This refers to the H-H equation (or the ICE Table approsch). (c). What is the conjugate base of HAe? What is the hydrolysis constant of that conjugate base (lt should not be called as the base's "dissociation constant")? Adding 0.40 g NaOH 100 mL of 2.0 M HAc resulted in a complete dissolution. 2. Given Ksp 6.0x10- for PbCh. Please calculate this ppt solubility (in molarity) in the cases below: (a) in pure water (b) in 0.1 M NaCl solution. (e) in 0.1 M Pb(NOs)h solution.

Explanation / Answer

1.

(a) HAc <==> H+ + Ac-

let x be the amount of dissociation

K = [Ac-][H+]/[HAc]

with,

[HAc] = 0.001 M

we get,

1.8 x 10^-5 = x^2/0.001

x = [H+] = 1.34 x 10^-4 M

The extent of reaction is amount of dissociation = 1.34 x 10^-4 x 100/0.001 = 13.4%

(b) Conjugate base of HAc is Ac-.

hydrolysis constant for Ac- would be,

Ac- + H2O <==> HAc + OH-

K(H) = [HAc][OH-]/[Ac-] = Kw/Ka

Addition of NaOH

moles of NaOH = 0.4 g/40 = 0.01 mols

moles of HAc = 2 M x 0.1 L = 0.2 mols

remaining [HAc] = (0.2 - 0.01)/0.1 = 1.9 M

[Ac-] = 0.01/0.1 = 0.1 M

pH = pKa + log([Ac-]/[HAc])

     = 4.74 + log(0.1/1.9)

     = 3.46

2. With Ksp for PbCl2 = 6 x 10^-5

(a) In pure water

let x amount of PbCl2 is in solution

Ksp = [Pb2+][Cl-]^2

6 x 10^-5 = (x)(2x)^2

x = 0.025 M is the solubility

(b) In 0.1 M NaCl

6 x 10^-5 = [Pb2+](0.1)^2

[Pb2+] = 6 x 10^-3 M is the solubility

(c) In 0.1 M Pb(NO3)2

6 x 10^-5 = (0.1).(2x)^2

x = 0.012 M is the solubility

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