Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74.
How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl?
c = 0.05 M
pKa = 4.74
pKa = -log (Ka)
Ka = 1.82 x 10-5
When HCl is added to the solution, the concentration of H+ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.
Case I: When 0.01 M HCl is taken.
Let x be the amount of acetic acid dissociated after the addition of HCl.
CH3COOH ↔ H+ + CH3COO-
Initial Conc. 0.05M 0 0
After dissociation 0.05-x 0.01+x x
As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 - x and 0.01 + x can be taken as 0.05 and 0.01 respectively.
Ka = [CH3COO-] [ H+] / [CH3COOH]
∴ Ka = (0.01) (x) / (0.05)
x = 1.82 x 10-5 x(multiply) 0.05 / 0.01
x = 1.82 x 10-3 x(multiply) 0.05
Now
α = Amount of acid dissociated / amount of acid taken
= 1.82 x 10-3 x 0.05 / 0.05
= 1.82 x 10-3
Case II: When 0.1 M HCl is taken.
Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:
[CH3COOH] = 0.05 - X : 0.05M
[CH3COO-] = X
[ H+] = 0.1+X ; 0.1M
Ka = [CH3COO-] [ H+] / [CH3COOH]
∴ Ka = (0.1) (X) / (0.05)
x = 1.82 x 10-5 x(multiply) 0.05 / 0.1
x = 1.82 x 10-4 x(multiply) 0.05
Now
α = Amount of acid dissociated / amount of acid taken
= 1.82 x 10-4 x 0.05 / 0.05
= 1.82 x 10-4
The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1.
Balance the following redox reactions by ion – electron method :
(a) MnO4 – (aq) + I – (aq) → MnO2 (s) + I2(s) (in basic medium)
(b) MnO4 – (aq) + SO2 (g) → Mn2+ (aq) + HSO4– (aq) (in acidic solution)
(c) H2O2 (aq) + Fe 2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)
In a process, 701 J of heat is absorbed by a system and 394 J ofwork is done by the system. What is the change in internal energy for the process?
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?
In a reaction A + B2 → AB2 Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Which one of the following will have largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl2(g)
Density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be its density at STP?
The equilibrium constant for a reaction is 10. What will be the value of ΔG0 ? R = 8.314 JK–1 mol–1, T = 300 K.
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
At 700 K, equilibrium constant for the reaction:
H2 (g) + I2 (g) ↔ 2HI (g)
is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?
Define octet rule. Write its significance and limitations.
What do you understand by
(a) inert pair effect
(b) allotropy and
(c) catenation?
Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
(a) CH3COOH + HO- → CH3COO- + H2O
(b) CH3COCH3 + C-N → (CH3)2 C (CN) (OH)
(c) C6H5 + CH3C+O → C6H5COCH3
Why is Li2CO3 decomposed at a lower temperature whereas Na2CO3 at higher temperature?
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?
(a) KI3
(b) H2S4O6
(c) Fe3O4
(d) CH3CH2OH
(e) CH3COOH
The size of isoelectronic species — F–, Ne and Na+ is affected by
(a) Nuclear charge (Z )
(b) Valence principal quantum number (n)
(c) Electron-electron interaction in the outer orbitals
(d) None of the factors because their size is the same.
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
CH3COOH (l) + C2H5OH (l) ↔ CH3COOC2H5 (l) + H2O (l)
(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?