0.0625mol of HI(g) was placed in a 0.500L flask and allowed to reach equilibrium

Question

0.0625mol of HI(g) was placed in a 0.500L flask and allowed to reach equilibrium. Calculate the equlibrium concentration of HI if the equilibrium constant is 1.0 x 101. H2(g) + I2(g) 2HI(g) N2(g) + O2(g) 2NO(g) 0.75M N2 and 0.50M O2 is allowed to reach equilibrium. What is the equilibrium concentration of NO if the Keq = 1.0 x 10-1? A. 0.083 B. 0.17 C. 0.42 If the ion product of barium sulfate is 1.0 x 10-9 a precipitate will form True False The pH of a solution which is 0.01M HC6H7O6 and 0.1M NaC6H7O6 is: A. 4.10 B. 5.10 C. 8.90 Pleas show work. Thank you very much. I will rate:)

Explanation / Answer

1) 0.0625mol of HI(g) was placed in a 0.500L flask and allowed to reach equilibrium. Calculate the equlibrium concentration of HI if the equilibrium constant is 1.0 x 101. H2(g) + I2(g) 2HI(g)

Solution :-

Initial molarity of HI = 0.0625 mol / 0.500 L = 0.125 M

2HI(g) H2(g) + I2(g)

0.125            0          0

-2x                x           x

0.125-2x         x        x

1/Kc = [H2][I2]/[HI]^2

1/1.0*10^1 = [x][x]/[0.125-2x]^2

Solving for x we get

X= 0.0242 M

Therefore at equilibrium concentration of the HI = 0.125-2x = 0.125-(2*0.0242)= 0.078 M

2) N2(g) + O2(g) 2NO(g) 0.75M N2 and 0.50M O2 is allowed to reach equilibrium. What is the equilibrium concentration of NO if the Keq = 1.0 x 10-1? A. 0.083 B. 0.17 C. 0.42

Solution :-

N2(g) + O2(g) 2NO(g)

0.75       0.50            0

-x            -x               +2x

0.75-x    0.50-x         2x

Kc =[NO]^2/[N2][O2]

1.0*10^-1 = [2x]^2/[0.75-x][0.50-x]

0.1=4x^2 /[0.75-x][0.50-x]

Solving for x using the quadratic formula we get

X=0.0833

Therefore at the equilibrium concentration of NO = 2x = 2*0.0833 = 0.17 M

3) If the ion product of barium sulfate is 1.0 x 10-9 a precipitate will form True False

Solution :- ion product constant that is ksp is 1.0*10^-9 is very very small therefore precipitate will form So it is true.

4) The pH of a solution which is 0.01M HC6H7O6 and 0.1M NaC6H7O6 is: A. 4.10 B. 5.10 C. 8.90

Solution :-

Acid and its conjugate base makes the buffer solution therefore using the Henderson Hassel Balch equation we can calculate the pH

Pka of ascorbic acid = 4.2

pH= pka + log [conj.base]/[acid]

pH= 4.1 + log ([0.1]/[0.01])

pH= 5.1

So the pH of the solution is 5.1

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