This set of Biochemistry Interview Questions and Answers focuses on “Ionization of Water, Weak acids and Weak Bases”.
1. What is the concentration of OH– in a solution with a H+ concentration of 1.3×10-4 M?
a) 7.7×10-10 M
b) 7.7×10-9 M
c) 7.7×10-11 M
d) 7.7×10-12 M
Explanation: Kw = [H+ ][OH–] Solving for [OH–] gives
[OH– ]=(Kw )/([H+]) = (1.0 X 10-14)/(1.3 X 10-4 ) M
= 7.7 × 10-11 M.
2. Identify the triprotic acid from the following
a) Carbonic acid
d) Phosphoric acid
Explanation: Phosphoric acid is triprotic as it can give up three protons
3. Which of the following acids has the lowest pKa value?
a) Acetic acid
b) Sulphuric acid
d) Oxalic acid
Explanation: The stronger the acid, the lower its pKa. Sulphuric acid is the strongest acid of all the other acids mentioned, so its pKa is the least.
4. Which of the following bases has the highest pKa value?
Explanation: The stronger the base, the higher its pKa. NaOH is the strongest base of all the other bases mentioned, so its pKa is the highest.
5. The degree of ionization does not depend on?
c) Nature of solvent
Explanation: Degree of ionization is independent of current. It depends only on temperature, nature of solvent and concentration.
6. The hydrolysis constant of CH3 COONa is given by
a) Kh = Kw/Ka
b) Kh = Kw/Kb
c) Kh = Kw/Ka×Kb
d) Kh = Ka + Kb
Explanation: CH3 COONa is a salt of weak acid and strong base.
7. K1 and K2 for oxalic acid are 6.5×10-2and 6.1×10-5respectively. What will be the [OH–] in a 0.01M solution of sodium oxalate?
Explanation: The hydrolysis of C2 O42- is as follows
C2 O42-+ H2 O → HC2O4- + OH–
8. If pKb for fluoride at 25°c is 10.83, the ionization constant of hydrofluoric acid in water at this temperature is
Explanation: Kw = Ka × Kb
Ka = Kw / Kb
Ka = 10-14/-log (10.83) = 6.75 × 10-4.
9. Aqueous solution of the detergents are
Explanation: Detergent is a salt of weak acid and strong base.
10. If pH of solution of NaOH is 12.0 the pH of H2 SO4 solution of same molarity will be
Explanation: pH = 12 then pOH = 2
[OH–] = 10-2
Molarity of NaOH = 10-2
For H2 SO4, molarity = 10-2
[H+] = 2 × 10-2
pH = 2 – log2 = 1.7.
Sanfoundry Global Education & Learning Series – Biochemistry.
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