What is the pH of a soln which is 0.24 M NH_3 and 0.20 M NH_3Cl? (k_b = 1.8 time

Question

What is the pH of a soln which is 0.24 M NH_3 and 0.20 M NH_3Cl? (k_b = 1.8 times 10^-5) a) 10.05 b) 9.34 c) 11.9 d) 11.1 Calculate the pH of a 0.15 M NH_3 solution. 1.8 times 10^-5 a) 3.5 b) 11.2 c) 2.8 d) 8.6 The pH of a solution formed when 45 mL of 0.1 M NaOH is added to 50 mL of 0.1 M CH_3COOH(K_a = 1 1.8 times 10^-5) is a) 5.7 b) 1.8 c) 9.2 d) 11.3 Calculate the molar solubility of CaF_2 at 25 degree C(K_sp-5.3 times l0^-9) in 0.01 M Ca(NO_3)_2)(aq) a)4.4 times 10^-6 b) 6.5 times 10^-21 c) 3.6 times 10^-4 d) 1.6 times 10^-7 A 1.00-L solution saturated at 25 degree C with calcium oxalate (CaC_2O_4) contains 0.0061 g of CaC_2O_4. The solubility constant (K_sp) for this salt is [CaC_2O_4 = 128.09 g/mol] a) 6 times 10^-3 b) 4.6 times 10^-8 c) 2.3 times 10^-9 d) 1.5 times 10^-4 Calculate the molar solubility of Ag_2SO_4 (K_sp = 1.4 times 10^-5) in 1.00 M Na_2SO_4 (aq) a) 0.0065 b) 0.0089 c) 0.0005 d) 0.0019 The value of Delta G degree_ f at 25 degree C for gaseous mercury is 31.85 KJ/mol. What is the vapor pressure of mercury at 25 degree C? a) 7.2 times 10^24 b) 2.6 times 10^-6 c) 9.1 times 10^-20 d) 6 times 10^-10 The eqm. Constant for a reaction is 2.48 at 25 degree C. What is the value of Delta G degree (kJ) at this temperature (R = 8.314 J/k. mol) a) 1.8 b) -1.8 c) -4.2 d) -2.25 For the Haber process for the synthesis of ammonia: N_2(g) + 3H_2 (g) = 2NH_3(g), what is Delta G (kJ/m) at 298 K for a reaction mixture that consists of 1.5 atm N_2, 3 atm H_2 and 1.8 atm NH_3? [Delta G degree (NH_3) = 16.66kJ/mol] a) -45 b) -11.6 c) -39.5 d) 22.4 At 25 degree C, the vapor pressure of water is 23.8 mmHg and enthalpy of vaporization is 44kJ/mol, the Clausius-Clapeyron eqn to calculate the vapor pressure (mmHg) of water at 40 degree C. a) 55.7 b) 10.1 c) 21.8 d) 92 CONSTANTS: Atomic weight: C = 12.01, H = 1.007, O = 15.99 N = 14; R = 0.0821 Latm/mol K = 8.314 J/mol K; 1 atm = 760 torr = 1.01325 times 10^5 Pa

Explanation / Answer

(11)

Formula to calculate, pOH of base buffers is,

pOH = pKb + Log[salt]/[base]

pOH = - Log(1.8 * 10-5) + Log(0.20 / 0.24)

pOH = 4.66

We know that, pH + POH = 14

pH = 14 - 4.66

pH = 9.33

(b)

(12)

accoding to Ostwald's theory of weak bses,

[OH-] = sqrt.(Kb*C) = sqrt.(1.8*10-5*0.15) = 1.64 * 10-3 M

pOH = -Log[OH-] = - Log(1.64 * 10-3) = 2.78

pH + pOH = 14

pH = 14 - 2.78

pH = 11.2

(b)

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