A) \[0.7051\]
B) \[0.6021\]
C) \[10.051\]
D) \[8.052\]
Correct Answer: B
Solution :
Total volume after mixing \[=50+30=80cc\] Molarity of \[HCl\] after mixing \[=\frac{50}{80}M\] Molarity of \[NaOH\] after mixing \[=\frac{30}{80}M\] Net molarity of \[HCl\] after mixing \[=\frac{50}{80}-\frac{30}{80}=0.25M\] \[[{{H}^{+}}]=0.25=2.5\times {{10}^{-1}}\] \[pH=-\log [{{H}^{+}}]\] \[=-\log [2.5\times {{10}^{-1}}]\] \[=1-0.3979=0.6021\]You need to login to perform this action.
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