A) 1
B) 13
C) 2
D) 0.1
Correct Answer: A
Solution :
\[[{{H}^{+}}\text{ }\!\!]\!\!\text{ }+\text{ }0.1/500/1000=\frac{0.1\times 1000}{500}=0.2\] \[pH=-\log \,[{{H}^{+}}]\] \[=-[\log 0.2]=-log\,[2\times {{10}^{-1}}]\] \[=-[\log 2-1\,\log \,10]\] \[=-\log 2+1\] = - 0.3010 + 1 = 0.6990 \[\approx \] 1.
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