A) 4.5
B) 4.0
C) 4.3
D) 3.3
Correct Answer: D
Solution :
[d] 3.3 In the final solution, \[[{{H}^{+}}]=({{10}^{-6}}+{{10}^{-3}})/2\] \[=\frac{{{10}^{-3}}({{10}^{-3}}+1)}{2}\] \[=\frac{1.00\times {{10}^{-3}}}{2}\] \[=0.5005\times {{10}^{-3}}\] \[=0.5005\times {{10}^{-4}}\] \[PH=-\log [{{H}^{+}}]\] \[=-\log (5.005\times {{10}^{-4}})\] \[=4-0.6991\] \[\simeq 3.3\]You need to login to perform this action.
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