A) 4
B) 5
C) 6
D) 7
Correct Answer: A
Solution :
Given,\[Ka=1.8\times {{10}^{-5}}\] [salt]\[=1\times 10=10\]millimol [Acid]\[=2\times 50=100\] millimol From Henderson's equation \[pH=pKa+\log \frac{[salt]}{[acid]}\] \[=-\log (1.8\times {{10}^{-5}})+\log \frac{10}{100}\] \[=-\log 1.8+5+\log {{10}^{-1}}\] \[=-0.2553+5-1=3.7447\] \[\approx 4\]You need to login to perform this action.
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