14. At-80°C, the equilibrium constant for the reaction NO is 4.66 x 10. At equil
ID: 1018435 • Letter: 1
Question
14. At-80°C, the equilibrium constant for the reaction NO is 4.66 x 10. At equilibrium, there was found to be 4.00 mole N,O, in a 1.0L flask. What is the equilibrium concentration of NO,? reaction N2o,-> 2N0210 a. 1.08 x 103 M b. 2.32 x 102 M c. 4.32 x10 M d. 1.17 x 10 M e. 5.68 x 10M 15. What is the equilibrium constant for the following reaction given is 5.3 x 10-10 that K, for HN, a. 5.3 x 1010 b. 1.9 x 10 C. 1.9 x 10 d. 5.3 x 10 e 1.9x10 At 0 °C, the ion product constant for water, K., is 1.2 x 10.15. The pH of water at 0°C is: 16. a. 7.46 b. 6.88 c. 7.56 d. 7.00 e. none of the above The equilibrium constants (K) for HCN and HF at 25°C are 6.2 x 10-10 and 7.2 x 10 17. respectively. The relative order of base strengths is: e. none of the aboveExplanation / Answer
14] N2O4 -----> 2NO2
k = [NO2]^2 / N2O4
[N2O4] = 4M
substitute k value and N2O4 concentration
[NO2] = 1.08 *10^-3 M
15] N3- + H2O -----> NH3 + OH-
Kc = [NH3] [OH-] / [N3-]
Ka of N3H = [ N3-] [H+] / [N3H]
Kc = Kw / Ka = 10^-14 / 5.3*10^-10 = 1.9 *10^-5
16] Kw = 1.2*10^-15
[H+] [OH-] = 1.2*10^-15
H+ and OH- are equal concentrations
[H+] = 3.46*10^-8
pH = 7.46
17] Higher Ka vaues of acids gives lower conjugate base
So
F- < H2O < CN-
18] pH = - log [H+]
[H+] = 10^-pH = 2*10^-12 M
19 ] pOH = - log [OH-] = 6.8
pH = 14 - pOH = 7.2
Answer none of the above
20] NO2-
21] [H+] = 10^-pH = 4.6*10^-9 M
22] pH + pOH = 14
pOH = 10.96
[OH-] = 1.096 *10^-11 M
23] SInce HNO3 completely dissociates
pH = -log [0.2 ] = 0.7
24] Ka = 4*10^-4 and 0.25M
HNO2 -----> H+ + NO2-
0.25 0 0
0.25-x x x
Ka = x^2 / 0.25-x
keeping all values x = 9.8*10^-3
pH = -logx = 2
25] Ka = x^2 / 17.6 -x
since x is very small
we can approximate 17.6-x as 17.6
x = 0.0177
pH = -logx = 1.75
26 ] [H+ ] = 10^-pH = 3.16*10^-3 = x^2
Ka = x^2 / y - x
keep Ka , x value and find y value
y = 0.156 is concentration of benzoic acid
concentration = moles / volume
moles = 0.156 *0.5 = 0.078
c is the answer
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