A) 4
B) 8
C) 4
D) 10
Correct Answer: D
Solution :
30 mL of \[\text{0}\text{.02}\,\text{M}\,\text{N}{{\text{H}}_{\text{4}}}\text{OH}\,\text{+}\,\text{15}\,\text{mL}\]of \[\text{0}\text{.02}\,\text{M}\,\text{HCl}\] Number of millimoles of \[\text{N}{{\text{H}}_{\text{4}}}\text{OH}\,\text{=}\,\text{30}\,\text{ }\!\!\times\!\!\text{ }\,\text{0}\text{.02}\] \[\text{=}\,\text{0}\text{.6}\,\text{mol}\] Number of millimoles of \[\text{HCl}\,\text{=15}\,\text{ }\!\!\times\!\!\text{ }\,\text{0}\text{.02}\,\text{=}\,\text{0}\text{.3}\,\text{mol}\] In an acid base reaction, salt will always form but we have to check what is left or consumed. \[HCl+N{{H}_{4}}OH\xrightarrow{{}}N{{H}_{4}}Cl+{{H}_{2}}O\] \[\begin{align} & \text{Initial}\,\,\,\,\,\text{0}\text{.3}\,\,\,\,\,\,\text{0}\text{.6}\,\,\,\,\,\,\text{0} \\ & \text{Final}\,\,\,\,\,\,\text{0}\,\,\,\,\,\,\,\,\,\,\,\text{0}\text{.3}\,\,\,\,\,\,\text{0}\text{.3} \\ \end{align}\] Basic buffer will form as weak base \[\text{N}{{\text{H}}_{\text{4}}}\text{OH}\]is left. \[[N{{H}_{4}}Cl]=\frac{\text{0}\text{.3}}{\text{Total}\,\text{volume}}=\frac{0.3}{45}\] \[[N{{H}_{4}}OH]=\frac{\text{0}\text{.3}}{\text{Total}\,\text{volume}}=\frac{0.3}{45}\] Applying \[pOH=p{{K}_{a}}+\log \frac{[Salt]}{[Base]}\] \[pOH=4+\log \frac{\frac{0.3}{45}}{\frac{0.3}{45}}\] \[[\because \,log\,1=0]\] \[pOH=4+\log \,1\] \[pOH=4\] \[\because \] \[pH=14-pOH\] \[\therefore \] \[pH=14-4\] \[pH=10\]
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