A) \[8.94\] and \[9.24\]
B) \[9.24\]and \[9.72\]
C) \[5.06\]and \[4.76\]
D) \[9.24\]and \[8.94\]
Correct Answer: A
Solution :
[a] It is a basic buffer. \[[Base]\,\,=\,\,[N{{H}_{3}}]\,\,=\,\,0.2M\] \[[Base]\,\,=\,\,[N{{H}_{3}}]\,\,=\,\,0.2M\] (?n? factor = 2) \[{{(N{{H}_{4}})}_{2}}S{{O}_{4}}\xrightarrow{{}}2\overset{\oplus }{\mathop{N}}\,{{H}_{4}}+S{{O}_{4}}^{2-}\] \[pOH=p{{K}_{b}}+\log \left[ \frac{salt}{base} \right]\] \[=4.76+\log \left( \frac{0.4}{0.2} \right)\] \[=4.76+0.3=5.06\] \[pH=14-5.06=8.94\] New \[pH:\,\,Ca{{(OH)}_{2}}\xrightarrow{{}}C{{a}^{2+}}+2\overset{\Theta }{\mathop{O}}\,H\] \[\left[ \overset{\Theta }{\mathop{O}}\,H \right]=0.05\times 2=0.1N\] (?n? factor =2) BBB Rule:- On adding base, to basic buffer, concentration of base increases and salt decreases \[[N{{H}_{3}}]new\,=0.2+0.1=0.3N,\] \[[{{(N{{H}_{4}})}_{2}}S{{O}_{4}}]=0.4-0.1=0.3N\] \[\therefore \,\,\,pOH=4.76+\left( \frac{0.3}{0.3} \right)=4.76,\] \[pH=14-4.76=9.24\]
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