Citric acid , (COOH)CH2-C(OH)(COOH)-CH2(COOH) is a triprotic acid with three acidic protons. It can be represented as H3A
The equilibrium reaction are:
1) H3A -------- H2A- + H+ pKa1 = 3.14
2) H2A- --------- HA2- + H+ pKa2 = 4.76
3) HA2- -------- A3- + H+ pKa3 = 6.40
The given pH = 4.2 which is closest to pKa2. hence the two ions that will be present are: H2A- and HA2-
As per Henderson Hasselbach equation:
pH = pka + log [HA2-]/[H2A-]-----------(1)
[HA2-]/[H2A-] = 10^(pH-pKa) = 10^(4.2-4.76) = 0.275
i.e. [HA2-] = 0.275 [H2A-]-----------------(2)
It is given that: [HA2-] + [H2A-] = 50 mM ---------------(3)
substituting for HA2- IN eq(3) we get:
1.275 [H2A-] = 50 mM
[H2A-] = 50/1.275 = 39.22 mM
[HA2-] = 0.275(39.22) = 10.79 mM
Answer:
The answers to the questions are;
Part A, endothermic
Part B, above 100 °C
Part C below 100 °C
Part D, 100 °C.
Explanation:
Part A endothermic,
The vaporization of water requires the addition of the heat of vaporization for the given mass of water as well as the heat required to raise the water to the boiling point.
Part B, above 100 °C
The boiling or vaporization of water is a spontaneous process above the boiling point for water which is 100 °C point
Part C below 100 °C
It is a non-spontaneous process below the boiling point of water which is 100 °C
Part D, 100 °C
Water is in equilibrium with steam at 100 °C.
Answer:
15
I may be wrong worked it out in my head
The term for the outermost electrons is calence electrons