A) 4
B) 8
C) 4
D) 10
Correct Answer: D
Solution :
30 mL of\[0.02M\,N{{H}_{4}}OH+15mL\text{ }of\text{ }0.02\text{ }M\text{ }HCl\] Number of millimoles of \[N{{H}_{4}}OH=30\times 0.02\] \[=0.6\text{ }mol\] Number of millimoles of\[HCl=15\times 0.02=0.3\]mol In an acid base reaction, salt will always form but we have to check what is left or consumed. \[HCl+N{{H}_{4}}OH\xrightarrow{{}}N{{H}_{4}}Cl+{{H}_{2}}O\] Initial 0.3 0.6 0 Final 0 0.3 0.3 Basic buffer will form as weak base\[N{{H}_{4}}OH\]is left. \[[N{{H}_{4}}Cl]=\frac{0.3}{Total\text{ }volume}=\frac{0.3}{45}\] \[[N{{H}_{4}}OH]=\frac{0.3}{Total\text{ }volume}=\frac{0.3}{45}\] Applying\[pOH=p{{K}_{a}}+log\frac{[salt]}{[Base]}\] \[pOH=4+\log \frac{0.3}{\frac{45}{\frac{0.3}{45}}}\] \[pOH=4+\log 1\] \[[\because \log 1=0]\] \[pOH=4\] \[\because \] \[pH=14-pOH\] \[\therefore \] \[pH=14-4\] \[pH=10\]
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