Consider the following balanced equilibrium expression 2 CrO^2-_4 + rightharpoon

Question

Consider the following balanced equilibrium expression 2 CrO^2-_4 + rightharpoonoverleftharpoon Cr_2O^2-_7 + H_2O Which of the following statements is true. a) Dissolving Na_2CrO_4 in an acidic solution b) Addition of Cr_2O^2-_7 ion will shift the equilibrium position to the right c) Addition of acid to the equilibrium mixture will form more CrO^2-_4 ions d) Dissolving Na_2Cr_2O_7 in pure water will result in an acidic solution e) Addition of pure water will shift the equilibrium to the left Rank the following aqueous solutions from most acidic to least acidic, 1) 0.1 M H_3PO_4 (K_a1 = 7.1 times 10^-3) 2) 0.1 M NaOH 3) 1 M HCl 4) 0.2 M C_6H_5NH_3Cl(C_6H_5NH_2 K_b 1.12 times 10^-3) 5) 0.1 M Ba(OH)_2 6) 0.1 M HClO_4 a) 5 > 2 > 1 > 4 > 6 > 3 b) 1 > 2 > 3 > 4 > 5 > 6 c) 3 > 6 > 1 > 4 > 2 > 5 d) 3 > 6 > 4 > 1 > 2 > 5 e) 1 > 3 > 5 > 2 > 4 > 6

Explanation / Answer

Use the following relations:

pH = -log [H+]

pH + pOH = 14

Ka = [H+][A-]/[HA] (for a weak acid, HA)

Kb = [BH+][OH-]/[B] (for a weak base, B)

1) Write down the dissociation reaction:

H3PO4 (aq) <=====> H+ (aq) + H2PO4- (aq)

   ---                             x                  x

Ka = [H+][H2PO4-]/[H3PO4] = (x).(x)/(0.1 – x)

===> 7.1*10-3 = x2/(0.1 – x)

Since Ka is small, assume small dissociation, i.e, x << 0.1 and hence (0.1 – x) 0.1. Therefore,

7.1*10-3 = x2/0.1

===> x2 = 7.1*10-4

===> x = 0.02664

Therefore, [H+] = 0.02664 M and pH = -log [H+] = -log (0.02664) = 1.5745 1.57.

2) Write down the dissociation of NaOH as below:

NaOH (aq) ------> Na+ (aq) + OH- (aq)

Due to the 1:1 nature of NaOH, we have [OH-] = 0.1 M; therefore, pOH = -log [OH-] = -log (0.1) = 1.0

Therefore, pH = 14 – pOH = 14 – 1.0 = 13 (ans)

3) Write down the dissociation of HCl as below:

HCl (aq) -------> H+ (aq) + Cl- (aq)

Due to the 1:1 nature of dissociation, [H+] = 1.0 M; therefore, pH = -log [H+] = -log (1.0) = 0 (ans).

4) Write down the reaction of C6H5HN3Cl (concerned species is C6H5NH3+) as below:

C6H5NH3+ (aq) -------> H+ (aq) + C6H5NH2 (aq)

Given Kb, Ka = Kw/Kb = (1.0*10-14)/(1.12*10-3) = 8.928*10-12

Ka = [H+][C6H5NH2][C6H5NH3+] = (x).(x)/(0.2 – x)

Since Ka is small, we can assume (0.2 – x) 0.2. Therefore,

8.928*10-12 = x2/0.2

===> x2 = 1.7856*10-12

===> x = 1.3363*10-6

Thus, [H+] = 1.3363*10-6 M and hence, pH = -log [H+] = -log (1.3363*10-6) = 5.874 5.87 (ans).

5) Write down the dissociation of Ba(OH)2 as below:

Ba(OH)2 (aq) <=====> Ba2+ (aq) + 2 OH- (aq)

Due to the 1:2 nature of Ba(OH)2, [OH-] = 2*0.1 M = 0.2 M; therefore, pOH = -log [OH-] = -log (0.2) = 0.6989.

Thus, pH = 14 – pOH = 14 – 0.6989 = 13.3011 13.30 (ans).

6) Write down the dissociation as

HClO4 (aq) ------> H+ (aq) + ClO4- (aq)

[H+] = 0.1 M; pH = -log (0.1) = 1.0 (ans)

The lower the pH, the more acidic is the solution. Hence, the order of acidity is

3 > 6 > 1 > 4 > 2 > 5 (ans)

Ans: (c) 3 > 6 > 1 > 4 > 2 > 5

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.