Equilibrium
Get insights from 241 questions on Equilibrium, answered by students, alumni, and experts. You may also ask and answer any question you like about Equilibrium
- 241
Questions
- 0
Discussions
- 0
Active Users
- 0
Followers
Have a question about Equilibrium? Ask our experts
Keep it short & simple. Type complete word. Avoid abusive language.
7.70. The ionization constant of benzoic acid is 6.46×10−5 and Ksp for silver benzoate is 2.5×10−13. How many times is silver benzoate more soluble in a buffer of pH= 3.19 compared to its solubility in pure water?
7.70. The ionization constant of benzoic acid is 6.46×10−5 and Ksp for silver benzoate is 2.5×10−13. How many times is silver benzoate more soluble in a buffer of pH= 3.19 compared to its solubility in pure water?
Since pH=3.19,
[H3O+]=6.46×10−4M
C6H5COOH+H2O↔C6H5COO−+H3O
Ka= [C6H5COO−] [H3O+] / [C6H5COOH]
[C6H5COOH] / [C6H5COO−]= [H3O+] / Ka=6.46×10−4 / 6.46×10−5=10
Let the solubility of C6H5COOAg be xmol/L.
Then,
[Ag+]=x
[C6H5COOH]+ [C6H5COO−]=x
10 [C6H5COO−]+ [C6H5COO−]=x
[C6H5CO
Upvote 
7.69. Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate, Ksp = 7.4 × 10–8).
7.69. Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate, Ksp = 7.4 × 10–8).
After mixing, the concentration of NaIO3? is 0.002 / 2? =0.001M.
After mixing, the concentration of Cu (ClO3? )2? is 0.002? / 2 =0.001M.
NaIO3? Na++IO3−?
[IO3−? ]=0.001M
Cu (ClO3? )2? Cu2++2ClO3−?
[Cu2+]=0.001M
The ionic product of Cu (IO3? )2? is
[Cu2+] [I−]
Upvote 7.68. The solubility product constant of Ag2CrO4 and AgBr are 1.1×10−12and 5.0×10-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
7.68. The solubility product constant of Ag2CrO4 and AgBr are 1.1×10−12and 5.0×10-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Silver chromate: Ag2CrO4? ? 2Ag+ + CrO42−?
[Ag+] = 2s1? , CrO42−? = s1?
Ksp? = (2s1? )2 (s1? ) = 4s3 = 1.1×10−12
s1? = 6.5 × 10−5 . (1)
Silver bromide: AgBr? Ag+ + Br−
[Ag+] = [Br−] = s2?
Ksp? = (s2? ) × (s2? ) = (s2)2? = 5.0 × 10−13
s2? =7.07×10−7. (2)
Divide equation (1) by equation
Upvote 7.67. Determine the solubility of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.
7.67. Determine the solubility of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.
1. Silver chromate:
Ag2CrO4→2Ag++CrO42−
Then, [Ag+] = (2s), [CrO42−] = s
Ksp= [Ag+]2 [CrO42−]
s = 0.65×10−4M
So, [Ag+] = 2s = 1.30 x 10−4M, [CrO42−] = 6.5 x 10-5M
2. Barium Chromate:
BaCrO4→Ba2++CrO42−
[Ba2+] = [CrO42−] = s
Then, Ksp= [Ba2+] [CrO42−] = s x s
= > 1.2 x 10-10M = s2
= > s = 1.09 x 10-5M
3. Ferric
Upvote 7.66. Calculate the pH of the resultant mixtures:
a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl
b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2
c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH
7.66. Calculate the pH of the resultant mixtures:
a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl
b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2
c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH
(a) Total number of moles present in 10 mL of 0.2 M calcium hydroxide are (10×0.2) / 1000? = 0.002 moles.
Total number of moles present in 25 mL of 0.1 M HCl are (25×2) / 1000? = 0.0025 moles.
Ca (OH)2? + 2HCl → CaCl2? + 2H2? O
1 mole of calcium hydroxide reacts with 2 moles of HCl.
0.0025
Upvote 7.65. Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?
7.65. Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?
Ionic product,
Kw= [H+] [OH−]
Let [H+]= x
Since [H+]= [OH−], Kw=x2.
⇒Kw at 310K is 2.7×10−14
∴2.7×10−14=x2
⇒x=1.64×10−7
⇒ [H+]=1.64×10−7
⇒ pH= −log [H+]=−log [1.64×10−7]=6.78
Hence, the pH of neutral water is 6.78.
Upvote 7.64. The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?
7.64. The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?
Given Ka=1.35 x 10−3
For acid solution:
[H+] = (KaC)1/2 = (0.00135 x 0.1)1/2 = 0.0116M
pH = – log [H+] = –log0.0116 = 1.936
For 0.1M sodium salt solution
ClCH2COONa is the salt of a weak acid i.e., ClCH2COOH and a strong base i.e., NaOH.
pKa = -logKa = log (0.00135) = 2.8697
pKw = 14
logc = lo
Upvote 7.63. Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH4NO3, NaNO2 and
7.63. Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH4NO3, NaNO2 and
(i) NaCl:
NaCl+H2O↔NaOH+HCl
Strong base Strong acid
Therefore, it is a neutral solution.
(ii) KBr:
KBr+H2O↔KOH+HBr
Strong base Strong acid
Therefore, it is a neutral solution.
(iii) NaCN:
NaCN+H2O↔HCN+NaOH
Weak acid Strong base
Therefore, it is a basic solution.
(iv) NH4NO3
NH4N
Upvote 7.62. A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
7.62. A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
pH=3.44
We know that,
pH=−log [H+]
∴ [H+]=3.63×10−4
Then, Kb= (3.63×10−4)2 / 0.02 (? concentration =0.02M)
⇒Kb=6.6×10−6
Now, Kb=Kw / Ka
⇒Ka=Kw / Kb=10−14 / 6.6×10−6=1.51×10−9
Upvote 7.61. The ionization constant of nitrous acid is 4.5×10−4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
7.61. The ionization constant of nitrous acid is 4.5×10−4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Since NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2);
NO2− + H2O ↔ HNO2 + OH−
Then, Kb= [HNO2] [OH−] / [NO2−]
⇒ Kw/ Ka = (10−14) / (4.5×10−4)
= 0.22×10−10
Now, if x moles of the salt undergo hydrolysis, and then the concentration of various species present in the solution will be:
Upvote