A) 0.187
B) 0.115
C) 0.076
D) 0.896
Correct Answer: B
Solution :
[b] Buffer capacity \[=\frac{No.of\,moles\,of\,base\,added/litere\,of\,buffer}{Change\,in\,pH}\] \[=\frac{0.01}{(6.832-6.745)}=\frac{0.01}{0.087}=0.11\]You need to login to perform this action.
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