A) 3.979
B) 0.6021
C) 12.042
D) 1.2042
E) 0.3979
Correct Answer: B
Solution :
Concentration of mixture \[cone\text{ }of\text{ }acid\times its\text{ }volume\] \[=\frac{-\,cone\text{ }of\text{ }base\times its\text{ }volume}{Total\text{ }volume}\] \[N=\frac{{{N}_{1}}{{V}_{1}}-{{N}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}\] \[=\frac{50\times 1-30\times 1}{50+30}\] \[=\frac{20}{80}\] \[=\frac{1}{4}\] \[=0.25\] Since\[HCl\]is in excess, so mixture is acidic and mixture contains \[[{{H}^{+}}]\]ion. Thus, \[[{{H}^{+}}]=0.25\] \[pH=-\log [{{H}^{+}}]\] \[=-log\text{ }0.25\] \[=0.6021\]
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