- 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
217 - Chemistry Questions Answers
A 0.1N solution of sodium bicarbonate has a pH value of
Please submit complete question, Ka should be given with percentage of ionisation
pH of 0.1 M NH4Cl solution is
For calculating pH of a salt, only concentration is not sufficient
pH of 102 M HCl is
This is a strong acid so consider 100 percent ionisation
we get [H+] = 102
it is more than 1 so pH = 0
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