A) 5.2
B) 9.4
C) 9.0
D) 5.0
Correct Answer: C
Solution :
\[{{H}_{2}}S{{O}_{4}}+2N{{H}_{4}}OH~\,\,\to \,\,{{(N{{H}_{4}})}_{2}}S{{O}_{4}}\,+\,2{{H}_{2}}O\] \[t\,\,=\,\,0\] 2 millmole 6 millimole \[t\text{ }=\text{ }t\] 0 2 millimole 2 millimole \[\therefore \,\,\,\,\,\,\,\,{{P}^{OH}}\,=\,\,{{P}^{kb}}\,\,+\,\,\log \,\,\frac{[Salt]}{[Base]}\] \[{{P}^{OH}}\,=\,\,4.7\,\,+\,\,\log \,\,\frac{4/50}{2/50}\] \[\therefore \,\,{{P}^{OH}}\,=\,4.7\,\,+\,\,0.3\,\,=\,\,5\] \[\therefore \text{ }\,{{P}^{H}}\text{ }=\text{ }9.\]You need to login to perform this action.
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