A) 2.5
B) 3.5
C) 5.5
D) 6.5
Correct Answer: B
Solution :
\[\because \] pH of HCl solution =3.0 \[\therefore \,[{{H}^{+}}]\] in HCl solution \[=1\times {{10}^{-3}}\] \[\therefore \] pH of NaOH solution =10.0 \[\therefore \,\,[{{H}^{+}}]\] in NaOH solution \[=1\times {{10}^{-10}}\] \[[O{{H}^{-}}]\] in NaOH solution \[=\frac{1\times {{10}^{-14}}}{1\times {{10}^{-10}}}\] Mill equivalents of \[HCl={{N}_{1}}{{V}_{1}}=2.0\times 1\times {{10}^{-3}}\] \[=2.0\times {{10}^{-3}}\] Mill equivalents of NaOH \[=3.0\times 1\times {{10}^{-4}}\] \[=3.0\times {{10}^{-4}}\] Since, mill equivalents of HCl are in excess, the mill equivalents of \[[{{H}^{+}}]\] in mixture \[=(2.0\times {{10}^{-3}})-(3.0\times {{10}^{-4}})\] \[=1.7\times {{10}^{-3}}\] Concentration of \[[{{H}^{+}}]\] in mixture \[=\frac{1.7\times {{10}^{-3}}}{3+2}=3.4\times {{10}^{-4}}\] pH of mixture \[=-\log [{{H}^{+}}]\] \[=-\log \,(3.4\times {{10}^{-4}})\] = 3.5
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