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Question: What is the pH of a 0.49 – M NaClO solution?...

What is the pH of a 0.49 – M NaClO solution?

Explanation

Solution

The pH of any solution tells the acidity and basicity of the solution. Lower pH indicates an acidic solution, while a higher pH indicates a basic solution. A pH has 14 scales. pH is calculated as the negative logarithm of hydrogen ion concentration. On the other hand, pOH is calculated as the negative logarithm of hydroxide ion concentration.
Formula used:
pH = 14 + pOH

Complete answer:
We have been given sodium hypochlorite that dissociates in water to form sodium ion and hypochlorite ion. The hypochlorite ions act as a weak base in the solution. The concentration of these ions is given as 0.49 M. This means, [ClO−]=[NaClO]=0.49M[Cl{{O}^{-}}]=[NaClO]=0.49M
The hypochlorite ion acts as weak base and react with hydrogen to form acid called hypochlorous acid and hydroxide ions as, HClO−(aq)+H2O(l)⇌HClO(aq)+OH−(aq)HCl{{O}^{-}}(aq)+{{H}_{2}}O(l)\rightleftharpoons HClO(aq)+O{{H}^{-}}(aq), which suggests that the pH will be less than 7, due to the acid formed.
As we know the base dissociation constant tells us the concentration of the hydroxide ion, it is written as, Kb=[HClO].[OH−][ClO−]{{K}_{b}}=\dfrac{[HClO].[O{{H}^{-}}]}{[Cl{{O}^{-}}]}
Suppose the number of ions of hypochlorous acid and hydroxide is xM, then for hypochlorite ion, the concentration is (0.49 – x) M
So, putting these in the base dissociation formula, we have, Kb=x.x0.49−x=x20.49−x{{K}_{b}}=\dfrac{x.x}{0.49-x}=\dfrac{{{x}^{2}}}{0.49-x}as the value of x is small so the hypochlorite concentration is ≃\simeq 0.49, this implies that Kb=x20.49{{K}_{b}}=\dfrac{{{x}^{2}}}{0.49}
Therefore, the concentration of hydroxide can be written as, x=0.49×Kbx=\sqrt{0.49\times {{K}_{b}}} = [OH−]=0.49.KbM[O{{H}^{-}}]=\sqrt{0.49.{{K}_{b}}}M
As, we know pOH is negative log of hydroxide ion concentration, so we have,
pOH=−loh([OH−])pOH=-loh([O{{H}^{-}}]) , putting this into the pH formula we have,
pH=14+log⁡([OH−])pH=14+\log ([O{{H}^{-}}])
So, pH=14+log⁡(0.49.Kb)pH=14+\log (\sqrt{0.49.{{K}_{b}}})
Hence, the pH of the solution is pH=14+log⁡(0.49.Kb)pH=14+\log (\sqrt{0.49.{{K}_{b}}}).

Note:
We can take out the numerical value by putting the value of base dissociation constant. Dissociation constant tells us the extent of dissociation of any acid or a base. The relation between dissociation constant of an acid and base is, Ka.Kb=1.0×10−14{{K}_{a}}.{{K}_{b}}=1.0\times {{10}^{-14}} , which can also be used to determine the pH if we have the value of dissociation constant of acid.