| Determine pH of a 0.01 M aqueous solution of \[Cl{{C}_{6}}{{H}_{4}}N{{H}_{3}}Cl\]. |
| \[[{{K}_{b}}(Cl{{C}_{6}}{{H}_{4}}N{{H}_{3}})=4\times {{10}^{-13}}\], \[\log 19=1.3,\,\,log2=0.3,\,\,\sqrt{16.25}=4.02]\] |
A) 2.1
B) 2.5
C) 3.5
D) 3.1
Correct Answer: A
Solution :
| \[{{K}_{h}}=\frac{{{10}^{-14}}}{4\times {{10}^{-13}}}\] |
 |
| \[{{K}_{h}}=\frac{C{{\alpha }^{2}}}{(1-\alpha )}=0.025\Rightarrow \frac{0.01{{\alpha }^{2}}}{1-\alpha }=0.025.\] |
| \[\Rightarrow {{\alpha }^{2}}+2.5\alpha -2.5=0\] |
| \[\Rightarrow \] \[\alpha =0.76\] |
| \[[{{H}^{+}}]=7.6\times {{10}^{-3}}\Rightarrow pH=-\log [{{H}^{+}}]=2.1\] |
You need to login to perform this action.
You will be redirected in
3 sec
