Answer:
pKa of the acid HA with given equilibrium concentrations is 6.8
Explanation:
The dissolution reaction is:
HA ⇔ H⁺ + A⁻
So at equilibrium, Ka is calculated as below
Ka = [H⁺] x [A⁻] / [HA] = 2.00 x 10⁻⁴ x 2.00 x 10⁻⁴ / 0.260
= 15.38 x 10⁻⁸
Hence, by definition,
pKa = -log(Ka) = - log(15.38 x 10⁻⁸) = 6.813
The given question is incomplete . The complete question is :
In ionic bonding, during the transfer of electrons between two neutrally charged atoms, one electron moves from one atom to another. What are the new relative charges between the two atoms?
a. The giving atom and receiving atom are both negatively charged
b. The giving atom is now positively charged and the receiving atom is now negatively charged.
c. The giving and receiving atom are both positively charged
d. The giving atom is now negatively charged and the receiving atom is now positively charged.
Answer: The giving atom is now positively charged and the receiving atom is now negatively charged.
Explanation:
Ionic compounds are formed by transference of electrons between metals and non metals. The bond formed between a metal and a non-metal is always ionic in nature.
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Thus when one electron moves from one atom to another, the metal acquires a positive charge and the non metal acquires negative charge.
For example:
is formed by transfer of one electron from sodium to chlorine , thus forming
and 
<u>Ionic Bond</u> is formed when the electronegativity difference is 0.4 > 2.0. Electronegativity is a term that can be defined as a tendency of an atom to attract electron towards its own self.
Explanation:
Electronegativity is a term that can be defined as a tendency of an atom to attract electron towards its own self.
An electronegativity of an atom is affected by
- The atomic number of the atom
- Secondly by the distance at which the valence electron are residing from the nucleus
1. In case the electronegativity difference (which is denoted by ΔEN) is less than 0.5 then the bond formed is known as N<u>onpolar covalent.
</u>
2. In case the ΔEN is in between 0.5 and 1.6, the bond formed is referred to as the<u> Polar covalent
</u>
3. In case the ΔEN is more /greater than 2.0, then the bond formed is referred to as<u> Ionic Bond</u>
<u>2 Examples of Ionic bonds</u>
- The formation of sodium fluoride, NaF, from a sodium atom and a fluorine atom is an example of Ionic bond formation.
- Another example is the formation of NaCl from sodium (Na),which is a metal, and chloride (Cl), which is a nonmetal