A) 9.4
B) 5.02
C) 9.0
D) 5.2
Correct Answer: C
Solution :
20 ml 0.1 M \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]\[\Rightarrow \] \[{{\eta }_{{{H}^{\,+}}\,}}=\,\,4\] |
30 ml 0.2 M \[\text{N}{{\text{H}}_{\text{4}}}\text{OH}\]\[\Rightarrow \] \[{{\eta }_{N{{H}_{4}}OH}}=\,6\] |
\[\text{N}{{\text{H}}_{4}}\text{OH}+{{\text{H}}^{+}}\,\,\text{NH}_{4}^{\oplus }+{{\text{H}}_{2}}\text{O}\] |
\[\begin{align} & \Rightarrow 6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\ & \Rightarrow 6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\ \end{align}\] |
Solution is basic buffer |
\[\text{pOH}\,=\,\text{P}{{k}_{b}}\,+\text{log}\frac{\text{NH}_{\,4}^{\,+}}{\text{N}{{\text{H}}_{4}}\text{OH}}\] |
\[=\,4.7+\text{log}2\,=\,4.7+0.3\,=\,5\] |
\[\text{PH}\,=\,14\,-\,5\,=\,9\] |
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