- ✓$4$
- B$7$
- C$10$
- D$14$
${K_b} = \frac{{[O{H^ - }]\,\,[HX]}}{{[{X^ - }]}}$
$HX ⇌ {H^ + } + {X^ - }$
${K_a} = \frac{{[{H^ + }]\,\,[{X^ - }]}}{{[HX]}}$
$\therefore \,{K_a} \times {K_b} = [{H^ + }]\,\,[O{H^ - }] = {K_w} = {10^{ - 14}}$
Hence ${K_a} = {10^{ - 4}}$
Now as $[{X^ - }] = [HX],\,\,pH = p{K_a} = 4$.
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$(A)$ Atoms of different elements differ in mass.
$(B)$ Matter consists of divisible atoms.
$(C)$ Compounds are formed when atoms of different element combine in a fixed ratio.
$(D)$ All the atoms of given element have different properties including mass.
$(E)$ Chemical reactions involve reorganisation of atoms.
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