A) 13.4
B) 12.4
C) 7.6
D) 1.6
Correct Answer: B
Solution :
Let the volume of NaOH used = V = volume of HCl used. Concentration of mixture \[[O{{H}^{-}}]\] \[v=\frac{0.1\,\,V-0.05\,V}{(V+V)}\] [\[\because \] Milliequivalent of NaOH is in excess.] \[v=\frac{V(0.1\,-0.05\,)}{2V}\] \[=\frac{0.05}{2}=0.025\] \[pOH=-\log [O{{H}^{-}}]=-\log (0.025)\] = 1.602 \[\therefore \] pH = 14 - 1.602 = 12.4You need to login to perform this action.
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