Just 6, 7, 8 and 9 please
equilibrium? If not, in what direction will the reaction proceed? 3. For the reaction CO (8)+ Cl2 (8)COC2 (g) Ke-255 at 1000 K. If a reaction mixture initially contains 0.1500 M CO and 0.175 M Cl2 at 1000K, what are the equilibrium concentrations of all the reactants and products? Complete the table 4. H30\'1 ?? 3.5 x 103 1.8x109 For each strong acid solution, determine [H3O*], pH, and pOH 3.8x107 7.15 5. a. 0.25 M HCI b. 0.015 M HNO 6. Determine the [H30] & pH of an HNO2 solution of each concentration a. 0.500 M b. 0.100 M 7. For each strong base solution, determine [OH], pOH, and pH a. 0.15 M NaOH b. 1.5x103 M Ca(OH)2 8. Determine the [OH], pH, and pOH of a 0.125 M CO32 solution. 9. A 0.135 M solution of a weak base has a pH of 11.23. Determine Kb for the base.
Solution
6)
HNO2 ------> H+ + NO2-
pKa of HNO2 = 3.15
pH of a weak acid = 1/2(pKa - logC)
a)
pH = 1/2(3.15 - log(0.5))
pH = 1.7255
-log[H+] = 1.7255
[H+] = 10^-1.7255 = 0.0188
b)
pH = 1/2(3.15 - log(0.1))
pH = 2.075
-log[H+] = 2.075
[H+] = 10^-2.075 = 0.008414
7)
NaOH is Stron base so [OH-] = 0.15
pOH = -log[OH-]
pOH = -log(0.15) = 0.824
pH = 14 - pOH = 14 - 0.824 = 13.176
Ca(OH)2 is also a strong base so [OH-] = 1.5*10^-3
pOH = -log(1.5*10^-3) = 2.824
pH = 14 - pOH = 14 - 2.824 = 11.176
8)
CO32- is base coming from the acid H2CO3 so Kb = 5.6*10^-11
pKb of CO32- = -log(5.6*10^-11) = 10.252
pOH = 1/2(pKb - logC)
pOH = 1/2(10.252 - log(0.125))
pOH = 5.578
[OH-] = 10^-5.578 = 2.6424*10^-6
pH = 14 - 5.578 = 8.422
9)
pH = 11.23
pOH = 14 - pH = 14 - 11.23 = 2.77
pOH = 1/2(pKb - logC)
2.77 = 1/2(pKb - log(0.135))
pKb = 4.67
.
Russian Call Girls in Andheri Airport Mumbai WhatsApp 9167673311 💞 Full Nigh...
Equilibrium concentrations and acid-base calculations
1. Just 6, 7, 8 and 9 please
equilibrium? If not, in what direction will the reaction proceed? 3. For the reaction CO (8)+ Cl2
(8)COC2 (g) Ke-255 at 1000 K. If a reaction mixture initially contains 0.1500 M CO and 0.175
M Cl2 at 1000K, what are the equilibrium concentrations of all the reactants and products?
Complete the table 4. H30'1 ?? 3.5 x 103 1.8x109 For each strong acid solution, determine
[H3O*], pH, and pOH 3.8x107 7.15 5. a. 0.25 M HCI b. 0.015 M HNO 6. Determine the [H30]
& pH of an HNO2 solution of each concentration a. 0.500 M b. 0.100 M 7. For each strong base
solution, determine [OH], pOH, and pH a. 0.15 M NaOH b. 1.5x103 M Ca(OH)2 8. Determine
the [OH], pH, and pOH of a 0.125 M CO32 solution. 9. A 0.135 M solution of a weak base has a
pH of 11.23. Determine Kb for the base.
Solution
6)
HNO2 ------> H+ + NO2-
pKa of HNO2 = 3.15
pH of a weak acid = 1/2(pKa - logC)
a)
pH = 1/2(3.15 - log(0.5))
pH = 1.7255
-log[H+] = 1.7255
[H+] = 10^-1.7255 = 0.0188
b)
pH = 1/2(3.15 - log(0.1))
2. pH = 2.075
-log[H+] = 2.075
[H+] = 10^-2.075 = 0.008414
7)
NaOH is Stron base so [OH-] = 0.15
pOH = -log[OH-]
pOH = -log(0.15) = 0.824
pH = 14 - pOH = 14 - 0.824 = 13.176
Ca(OH)2 is also a strong base so [OH-] = 1.5*10^-3
pOH = -log(1.5*10^-3) = 2.824
pH = 14 - pOH = 14 - 2.824 = 11.176
8)
CO32- is base coming from the acid H2CO3 so Kb = 5.6*10^-11
pKb of CO32- = -log(5.6*10^-11) = 10.252
pOH = 1/2(pKb - logC)
pOH = 1/2(10.252 - log(0.125))
pOH = 5.578
[OH-] = 10^-5.578 = 2.6424*10^-6
pH = 14 - 5.578 = 8.422
9)
pH = 11.23
pOH = 14 - pH = 14 - 11.23 = 2.77
pOH = 1/2(pKb - logC)
![Just 6, 7, 8 and 9 please
equilibrium? If not, in what direction will the reaction proceed? 3. For the reaction CO (8)+ Cl2
(8)COC2 (g) Ke-255 at 1000 K. If a reaction mixture initially contains 0.1500 M CO and 0.175
M Cl2 at 1000K, what are the equilibrium concentrations of all the reactants and products?
Complete the table 4. H30'1 ?? 3.5 x 103 1.8x109 For each strong acid solution, determine
[H3O*], pH, and pOH 3.8x107 7.15 5. a. 0.25 M HCI b. 0.015 M HNO 6. Determine the [H30]
& pH of an HNO2 solution of each concentration a. 0.500 M b. 0.100 M 7. For each strong base
solution, determine [OH], pOH, and pH a. 0.15 M NaOH b. 1.5x103 M Ca(OH)2 8. Determine
the [OH], pH, and pOH of a 0.125 M CO32 solution. 9. A 0.135 M solution of a weak base has a
pH of 11.23. Determine Kb for the base.
Solution
6)
HNO2 ------> H+ + NO2-
pKa of HNO2 = 3.15
pH of a weak acid = 1/2(pKa - logC)
a)
pH = 1/2(3.15 - log(0.5))
pH = 1.7255
-log[H+] = 1.7255
[H+] = 10^-1.7255 = 0.0188
b)
pH = 1/2(3.15 - log(0.1))](https://image.slidesharecdn.com/just6-7-8and9pleaseequilibrium-ifnot-inwhatdirectionwill-230209170405-add5ebc6/85/Just-6-7-8-and-9-please-equilibrium-If-not-in-what-direction-will-docx-1-320.jpg)
![pH = 2.075
-log[H+] = 2.075
[H+] = 10^-2.075 = 0.008414
7)
NaOH is Stron base so [OH-] = 0.15
pOH = -log[OH-]
pOH = -log(0.15) = 0.824
pH = 14 - pOH = 14 - 0.824 = 13.176
Ca(OH)2 is also a strong base so [OH-] = 1.5*10^-3
pOH = -log(1.5*10^-3) = 2.824
pH = 14 - pOH = 14 - 2.824 = 11.176
8)
CO32- is base coming from the acid H2CO3 so Kb = 5.6*10^-11
pKb of CO32- = -log(5.6*10^-11) = 10.252
pOH = 1/2(pKb - logC)
pOH = 1/2(10.252 - log(0.125))
pOH = 5.578
[OH-] = 10^-5.578 = 2.6424*10^-6
pH = 14 - 5.578 = 8.422
9)
pH = 11.23
pOH = 14 - pH = 14 - 11.23 = 2.77
pOH = 1/2(pKb - logC)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)