A) 1.30
B) 4.2
C) 12.70
D) 11.70
Correct Answer: C
Solution :
\[50\,\,mL\]of \[0.1\,M\,HCl=\frac{0.1\times 50}{1000}=5\times {{10}^{-3}}\] \[50\,\,mL\]of \[0.2\,M\,NaOH=\frac{0.2\times 50}{1000}=10\times {{10}^{-3}}\] Hence, after neutralisation NaOH left. \[=10\times {{10}^{-3}}-5\times {{10}^{-3}}\] \[=5\times {{10}^{-3}}\] Total volume = 100 cc The concentration of NaOH \[=\frac{5\times {{10}^{-3}}\times 1000}{100}=0.05\,M\] \[[O{{H}^{-}}]=0.05\,M=5\times {{10}^{-2}}\,M\] \[pOH=-\,\log \,[O{{H}^{-}}]\] \[=-\,\log \,[5\times {{10}^{-2}}]\] \[=1.3010\] \[pH+pOH=14\] \[pH=14-1.3010\] \[=12.699=12.70\]
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