B. Consider the following at 25C, reaction When the initial to be 0.0130 M. of s

Question

B. Consider the following at 25C, reaction When the initial to be 0.0130 M. of so30g) 0.128 M concentration oogen gas at equilbriumis found a. Calculate the of Kc this reaction. for b. The value of Kp at 25 Cis what? c. a. What is the pH of a solution prepared by dissolving 050 g of Hclo. Mw.1onslin sufficient pure water to prepare 250.0 ml of solution? b. What is the pH of a solution prepared by dissolving 5.40 g of trimethylamine IN(CH) M.sot] sufficient pure water to prepare 250.0 ml of solution? in

Explanation / Answer

1. The reaction is 2SO3<--->2SO2+O2

[SO3] initially = 0.128M, let x= concentration of oxygen at equilibrium

So at equilibrium [SO3] =0.128-2x, SO2= 2x and O2= x =0.0130

hecne at equilibrium [SO3] =0.128-2*0.013= 0.102M, SO2= 2*0.013=0.026

KC= [SO2]2 [O2] /[SO3]2= 0.026*0.026* 0.013/(0.102)2= 0.00084

KP=Kc*(RT)deltan, deltan= moles of products-moles of reactants= 2+1-2= 1

Kp= 0.00084*(0.0821*298)1 = 0.02055

2. HClO4 is strong acid and ionizes completely

moles= mass/molar mass

molarity of 0.540 gm of HClO4 in 250ml (250/1000L= 0.25L) = moles/Liter of solution = 0.54/(100.5*0.25)=0.0215

HClO4+H2O ------->H3O++ ClO4-

[H3O+] = 0.0215,

pH= -log (0.0215)= 1.67

trimethyl amine is weak base, molarity = 5.4/59.11*0.25= 0.365M

(CH3)3N+ H2O ---------> (CH3)3NH+ + OH-

Kb= [ (CH3)3NH+] [OH-]/[(CH3)3N]

let x= drop in concentration of trimethyl amine to reach equilibrium

at equilibrium [(CH3)3NH+] =0.365-x, [(CH3)3NH+] = [OH-] =x

hence x2/(0365-x)= 7.4*10-5, when solved using excel, [OH- ] = 0.00515, [H+] = 10-14/0.00515=1.94*10-12

pH= 11.71

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

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