Q.1Explain why CH3COOH2^+ is in terms of protonic concept?
Q.2 0.1 mole HCl is dissolved in distilled water of volume V then at a limit V =infinite pH of the solution is equal to______?
Q.3One litre of water contains 10^-5 moles of H^+ ions at 25 degree Celcius.Degree of ionisation of water is ________?
Q.4 Approximate pH of 0.01M aqueous H2S solution when K1 and K2 for H2S at 25 degree Celcius are 10^-7 and 1.3*10^-13 respectively:______?
Q.1: To explain why CH3COOH2^+ is in terms of protonic concept, we need to understand the protonic concept of acids and bases.
In the protonic concept, acids are substances that can donate protons (H+ ions) and bases are substances that can accept protons.
CH3COOH2^+ is the conjugate acid of the acetate ion (CH3COO-). It can donate a proton to form the acetate ion. The chemical equation for this protonation reaction is:
CH3COOH + H+ -> CH3COOH2^+
Here, CH3COOH is the weak acid acetic acid and H+ is a proton. When acetic acid donates a proton, it becomes CH3COOH2^+, which is a positively charged species.
Therefore, CH3COOH2^+ can be explained as a species formed when acetic acid donates a proton, according to the protonic concept of acids and bases.
Q.2: To determine the pH of a solution when a certain amount of HCl is dissolved in distilled water, we need to understand the concept of pH and how it relates to the concentration of H+ ions.
pH is a measure of the acidity or alkalinity of a solution. It is defined as the negative logarithm (base 10) of the concentration of H+ ions in the solution.
In the given scenario, 0.1 mole of HCl is dissolved in distilled water. Since HCl is a strong acid, it completely dissociates in water to form H+ ions and Cl- ions. Therefore, the concentration of H+ ions is equal to the initial concentration of HCl.
Using the equation pH = -log[H+], we can calculate the pH of the solution. Since the concentration of H+ ions is 0.1 mole, the pH would be:
pH = -log(0.1) = 1
Therefore, when V (volume of water) approaches infinity, the pH of the solution would be 1.
Q.3: To determine the degree of ionization of water, we need to understand the concept of ionization and the dissociation of water.
Water (H2O) can undergo a process called self-ionization, where it dissociates into H+ ions and OH- ions:
H2O(l) ⇌ H+(aq) + OH-(aq)
The degree of ionization (α) represents the fraction of water molecules that have ionized into H+ and OH- ions.
In the given scenario, one liter of water contains 10^-5 moles of H+ ions. Since water is neutral, the concentration of H+ ions and OH- ions are equal. Therefore, the concentration of H+ ions and OH- ions is 10^-5 mole per liter.
The total concentration of water molecules is also one liter.
Using the equation for degree of ionization, α = (concentration of ions) / (total concentration of water molecules), we can calculate the degree of ionization:
α = (10^-5) / (1) = 10^-5
Therefore, the degree of ionization of water is 10^-5.
Q.4: To approximate the pH of a 0.01M aqueous H2S solution, we need to understand the equilibrium expression for the ionization of H2S and how it relates to the concentration of H+ ions.
The equilibrium equation for the ionization of H2S is:
H2S ⇌ H+ + HS-
This equilibrium can further ionize to form H+ and S2- ions:
HS- ⇌ H+ + S2-
The equilibrium constants for these reactions are denoted as K1 and K2, respectively.
In the given scenario, the concentration of H2S is 0.01M. Since H2S is a weak acid, it partially ionizes in water. To approximate the pH, we assume that the concentration of H+ ions contributed by the ionization of H2S is small compared to the initial concentration of H2S.
Using the equilibrium expression for the ionization of H2S, we have:
[H+] = √(K1 * initial concentration of H2S)
[H+] = √(10^-7 * 0.01) = 10^-5
Using the equation pH = -log[H+], we can calculate the approximate pH of the solution:
pH = -log(10^-5) = 5
Therefore, the approximate pH of the 0.01M aqueous H2S solution is 5.