Q G. Determine the number of moles of HI present when equilibrium is established

Question

Q G. Determine the number of moles of HI present when equilibrium is established in the reaction shown below if the mixture initially contains 0.0100 mol Ha and 0.0100 mol la. K 54.3 at 698 K) Q H. Calculate the pH of a 0.100 M solution of HCzHzFO2 (Ka = 2.6 x 10-3) QI A 25.00 mL of a 0.126 M solution of NaOH is titrated with a solution of HCl of unknown molarity. If the titration requires 28.3 mL of HCI, what is the molar concentration of HCI? QJ. What is the pH of a solution that is 0.100 M in NHs and 0.150 M in NHACI (Kb of ammot 1.8 x 105)

Explanation / Answer

1.For the reaction H2(g)+I2(g)<-> 2HI(g)

KC = [HI]2/{[H2]*[I2)}=54.3

Given initial concentration of H2 and I2 in 1 L= 0.01

Concentrations initially [H2] =[I2]=0.01/1=0.01 M

Let x= drop in concentration to reach equilibrium

At Equilibrium [H2] =[I2]=0.01-x and [HI]=2x

KC= (2x)2/(0.01-x)2= 54.3

Taking square root

2x/(0.01-x)=7.4

2x= 7.4*0.01-7.4x

9.4x= 7.4*0.01

x= 0.007872

since volume in 1 L

moles = concentration =x= 0.007872

at Equilibrium, moles of HI=2*0.007872=0.015744

2. HC2H2FO2+ H2O--------->C2H2FO2-+ H3O+

let x= drop in concentration of HC2H2FO2 to reach equilibirum

at Equilibrium [C2H2FO2-]= [H3O+]=x and [HC2H2FO2]=0.1-x

Ka= [C2H2FO2-] [H3O+]/[HC2H2FO2] = x2/(0.1-x)= 2.6*10-3

when solved using excel x= 0.0149, pH= -log [H3O+]= -log (0.0149)= 1.83

3. The reaction between HCl and NaOH is HCl+ NaOH--------->NaCl+ H2O

1 mole of HCl requires 1 mole of NaOH for neutralization

hence moles of HCl= moles of NaOH= molarity* volume in L, 1000ml= 1L

moles of NaOH= 0.126*25ml*(1/1000)L/ml = x*28.3/1000, x= molarity of HCl

x= 0.111

2. Henderson Hasselbacj equation is

pOH= pKb+ log [HB+]/[B],[ HB+]= conjugate acid in this case NH4+= 0.15M and [B]= concentration of NH3= 0.1

Kb= 1.8*10-5, pKb= -log Kb= 4.74

pOH= 4.74+log (0.15/0.1)= 4.92

pH= 14-pOH= 14-4.92= 9.08

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