Question

What is the ${{K}_{b}}$ of a weak base that produces an $O{{H}^{-}}$ per molecule if a 0.05M solution is 2.5% ionized?

Answer 1

Hint: The answer to this question is based on the formula to calculate the dissociation constant that is related to the degree of ionization which is given by ${{K}_{b}}=C\times {{\alpha }^{2}}$

Complete step by step solution:
In our classes in chemistry we have studied about the dissociation of base and acids in the physical chemistry part. Now, let us see the meaning of ionisation and how to relate this with the dissociation constant of base.
- Degree of ionisation is defined as the proportion of neutral particles which are ionized to the charged particles. Neutral particles may be gas or aqueous solution.
- The degree of ionisation depends upon the nature of solute, solvent and also temperature. Nature of the electrolyte plays an important role for the calculation of degree of ionisation.
- The ionisation of acids of bases depends on the solvent that is generally water. Weak acids ionise partially and strong acids ionise completely in water. Based on all the facts, the degree of ionisation of weak base is related to dissociation constant as, ${{K}_{b}}=C\times {{\alpha }^{2}}$where$\alpha $ is the degree of dissociation and C is concentration. Now, degree of ionisation according to the data can be written as,$\alpha =\dfrac{2.5}{100}=0.025$
C=0.05M.
Thus, substituting these values in the above equation we have,
 ${{K}_{b}}=0.05\times {{\left( 0.025 \right)}^{2}}=3.125\times {{10}^{-5}}$

Therefore, the correct answer is dissociation constant for a weak base that produces $O{{H}^{-}}$ per molecule if 0.05M solution is 2.5% ionised is $3.125\times {{10}^{-5}}$.

Note: Even in case of base, the one which ionise completely and produce $O{{H}^{-}}$ in aqueous solutions are the strong bases and one which ionises partially to produce $O{{H}^{-}}$ in aqueous solutions are the weak bases. Note that this point will lead to approaching the correct answer easily.