- Organic Chemistry
- Aldehydes and Ketones
- Alkyl Halides, Alcohols and ethers
- Amines and other nitrogen compounds
- Aromatic Chemistry
- Carbohydrates, Amino acids, protein, Vitamin and Fat
- Carboxylic acids and its derivatives
- Chemistry in daily life
- General Mechanism in organic compounds
- Hydrocarbons
- Nomenclature and isomerism
29 - Chemical and Ionic Equilibrium Questions Answers
100 c.c. of N/10 NaOH solution is mixed with 100 c.c. of N/5 HCl solution and the whole volume is made to 1 litre. The pH of the resulting solution will be
mili gm eq of NaOH = NV = 10
mili gm eq of HCl = 20
total mili gm eq = 20-10 = 10 of HCl
so N = 10/200 = 0.05
so pH = log[1/0.05]
When 0.1 mole of ammonia is dissolved in sufficient water to make 1 litre of solution. The solution is found to have a hydroxide ion concentration of 1.34 x 10-3 . The dissociation constant of ammonia is
NH3 + H2O---------------------> NH4+ + OH-
0.1 0 0
0.1(approx) 1.34*10-3 1.34*10-3
now calculate dissociation constant
Calculate the percentage ionization of 0.01 M acetic acid in 0.1 M HCl. Ka of acetic acid is 1.8 x 10-5
CH3COOH ----------------> CH3COO- + H+
0.01 0 0.1
0.01-x x 0.1+x
0.1+x =0.1 for x is negligibly small
so Ka = x*0.1/(0.01-x)
now solve
The dissociation constant of a weak acid HA and weak base BOH are 2 x 10-5 and 5 x 10-6 respectively. The equilibrium constant for the neutralization reaction of the two is (ignore hydrolysis of resuting salt)
HA + BOH ------------------> B+ + A- + H2O
K = [B+][A-][H2O]/[HA][BOH]
= [B+][OH-]/[BOH] * [H+][A-]/[HA] * [H2O]/[H+][OH-] = Kb * Ka / Kw
now solve
Which of the following increasing order of pH of 0.1 M solution of the component (a) HCOONH4 (b) CH3COOHNH4 (c) CH3COONa (d) NH4Cl is correct
(d) NH4Cl < (a) HCOONH4 < (b) CH3COONH4 < (c) CH3COONa
Which of the following molar ratio of NH3 and HCl in aqueous solution will constitute a buffer ? (a) 1 : 2 (c) 1 : 1 (b) 1 : 3 (d) 2 : 1
in (d) after reaction no. of moles of NH3 will be 1 and 1 mole NH4Cl will be formed so it will be a basic buffer.
In order to prepare a buffer of pH 8.26, the amount of (NH4 )2SO4 required to be mixed with 1L of 0.1M NH3 (pKb of NH3 = 4.74) is
NH3 + H2O ------------> NH4+ + OH-
so Kb = [NH4+] [OH-]/[NH3][H2O]
so Kb = [NH4+]/[NH3][H+] * [H+][OH-]/[H2O]
so Kb = [NH4+]/[NH3][H+] * Kw
now take log and solve. concentration of (NH4 )2SO4 will be double of [NH4+]
if you find any problem on solving this then reply
1.0 ml of dilute solution of NaOH is added to 100 ml of a buffer of pH 4. The pH of the resulting solution
(a) becomes 7.0
(b) becomes 9.0
(c) becomes 3.0
(d) Remains practically unchanged
it is strong acidic buffer so answer is (d)
10 ml of M/200 H2SO4 is mixed with 40 ml of M/200 H2SO4. The ph of the resulting solution is
M of H2SO4 = 1/200
so [H+] = 2*1/200 = 1/100 = 10-2
now calculate pH
The pH of solution at 25°C which has twice as many hydroxide ion as in pure water at 25°C , will be
no of hydroxide ion in pure water = 10-7
so according to the given condition no. of hydroxide ions in solution = 2*10-7
so pOH = -log2*10-7
so pH = 14 - pOH