A) \[-log\text{ }2\]
B) \[-log\text{ 0}.2\]
C) 1.0
D) 2.0
Correct Answer: C
Solution :
50 mL of \[0.4\text{ }NHCl=\frac{0.4}{1000}\times 50\,g\,eq.\] \[=0.02\text{ }g\text{ }eq.\] 50 mL of \[0.2\text{ }N\text{ }NaOH=\frac{0.2}{1000}\times 50\,g\,eq.\] \[=0.01\text{ }g\text{ }eq.\] After neutralisation,\[HCl\]left = 0.01 g eq. Total volume = 100 mL \[[HCl]=\frac{0.01}{100}\times 1000=0.1\,N\] \[=0.1\,M={{10}^{-1}}M\] \[pH=-\log [{{H}^{+}}]=-\log [{{10}^{-1}}]\] \[pH=1\]
You need to login to perform this action.
You will be redirected in
3 sec
