Sulphurous acid, H 2 SO 3 , is a weak acid capable of providing two H + ions. (a) What is the pH of a 0.45M solution of H 2 SO 3 ? (b) What is the equilibrium concentration of the sulphate ion SO 3 2− , in the 0.45M solution of H 2 SO 3 ?
Sulphurous acid, H 2 SO 3 , is a weak acid capable of providing two H + ions. (a) What is the pH of a 0.45M solution of H 2 SO 3 ? (b) What is the equilibrium concentration of the sulphate ion SO 3 2− , in the 0.45M solution of H 2 SO 3 ?
From the Ka1 and Ka2 values, Ka2 is smaller than the Ka1.
Therefore, H3O+ is almost produced entirely from Ka1.
Let’s calculate the H3O+ from Ka1.
First ionization:H2SO3(aq) + H2O(aq)⇌H3O+(aq) + HSO3-(aq)Equilibrium expression:Ka1 = [HSO3-][H3O+][H2SO3]Ka1= 1.2×10-2
Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.
H2SO3(aq) + H2O (aq)⇌ H3O+(aq) + HSO3-(aq)I 0.45M -- -- --C -x -- +x +xE (0.45-x) -- x x
The equilibrium concentration has to be determined for the sulphate ion SO32- in the 0.45M solution of H2SO3.
Concept Introduction:
Equilibrium constants:
The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.
Ka is acid constant for equilibrium reactions.
HA + H2O⇌H3O++ A-Ka= [H3O+][A-][HA]
Kb is base constant for equilibrium reaction.
BOH + H2O⇌B++ OH-Ka= [B+][OH-][BOH]
If a molecule donate only one hydrogen atom then the acid is called a monoprotic acid.
Example: HCl, HBr and HClO3
If each molecule can donate two or more hydrogen ions, the acid is a polyprotic acid.
A diprotic acid, such as sulfuric acid, H2SO4 has two acidic hydrogen atoms.
Some acids, such as phosphoric acid, H3PO4 are triprotic acids.
Polyprotic bases:
Can accept more than on proton (H+). So, it can react with water many times to produce many hydroxide ions (OH−).
Diprotic bases:
Can accept two protons.
Example: Sulphate ion SO4(aq)2-
Answer to Problem 73PS
The equilibrium concentration of the sulphate ion SO32- in the 0.45M solution of H2SO3 is 6.28×10-8M
Explanation of Solution
Second ionization:HSO3-(aq) + H2O(aq)⇌H3O+(aq) + SO32-(aq)Equilibrium expression:Ka2= [H3O+][SO32-][HSO3-]Ka2= 6.2×10-8
Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.
6.2×10-8 = y (6.77×10-2 + y)(6.77×10-2- y)Y is so small compared to 6.77×10-2So, all values are almost equal to 6.77×10-26.2×10-8 = y(6.77×10-2)(6.77×10-2) y = 6.2×10-8
Want to see more full solutions like this?
Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
The active ingredient of bleach such as Clorox is sodium hypochlorite (NaClO). Its conjugate acid, hypochlorous acid (HClO), has a Ka of 3.0 × 10–8.
(a)The undiluted bleach contains roughly 1 M NaClO. Calculate the pH of 1 M NaClO solution.
(b)Some applications require extremely diluted bleach solution, such as swimming pools. Suppose the solution in (a) is diluted by 10,000 -fold. Calculate the pH of the diluted solution, and demonstrate that you can still neglect the autoionization of water in your calculation.
(c)Suppose the solution in (a) is diluted by 1million-fold, briefly explain how your approach will be different. Write the equation with [H3O+] as the unknown, but you do not need to solve it.
Calculate the pH of 2.5 M CH2ClCOOH(aq) given that its Ka = 0.0014.
Chemists working with fluorine and its compounds some-
times find it helpful to think in terms of acid-base reac-
tions in which the fluoride ion (F¯) is donated and
ассеpted.
(a) Would the acid in this system be the fluoride donor or
fluoride acceptor?
(b) Identify the acid and base in each of these reactions:
CIF;O2 + BF;
CIF,O, · BF,
--
TiF, + 2 KF – K2[TiF,]
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.