A) Addition of 5 mL of 1 M HCl
B) Addition of 50 mL of 0.01 M HCl
C) Addition of 50 mL of 0.002 M HCl
D) Addition of Mg
Correct Answer: D
Solution :
Millimoles of 50 mL of 0.01 M HCl \[=50\times 0.01M=0.5\] pH of the solution \[=-\log (0.5)=0.3010\] = - log (0.5) = 0.3010 On adding 5 mL of 1 M HCl, Concentration of the resultant solution \[=\frac{50\times 0.01+5\times 1}{50+5}=0.1\] \[\therefore \]pH of the solution \[=-\log (0.1)=1\] On adding 50 mL of 0.01 M HCl, Concentration of the resultant solution \[=\frac{50\times 0.01+50\times 0.01}{50+50}=0.01\] \[\therefore \] pH of the solution \[=-\log (0.01)=2\] On adding 50 mL of 0.002 M HCl, concentration of the resultant solution \[=\frac{50\times 0.01+50\times 0.002}{50+50}=6\times {{10}^{-3}}\] \[\therefore \] pH of the solution \[=-\log (6\times {{10}^{-3}})=2.22\] On adding Mg, \[Mg+2HCl\xrightarrow{{}}MgC{{l}_{2}}+2{{H}^{+}}\] Since, \[pH\propto \frac{1}{[{{H}^{+}}]}\] \[\therefore \] In this case pH decreases.
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