A) 2
B) 1
C) 1.7
D) 0.3
Correct Answer: A
Solution :
Moles of \[{{H}_{2}}S{{O}_{4}}=\frac{0.49}{98}=5\times {{10}^{-3}}\] moles of \[{{H}_{2}}S{{O}_{4}}\] present per litre of solution (molarity) \[=\frac{.005}{1}=.005\]M. \[{{H}_{2}}S{{O}_{4}}+2{{H}_{2}}O\]⇌\[2{{H}_{3}}{{O}^{+}}+SO_{4}^{-\,-}\] one mole of \[{{H}_{2}}S{{O}_{4}}\] give 2 moles of \[{{H}_{3}}{{O}^{+}}\]ions. \[{{H}_{3}}{{O}^{+}}=2\times ({{H}_{2}}S{{O}_{4}})\]\[=2\times 0.005=0.01\]M \[[{{H}^{+}}]={{10}^{-2}}\]M ; \[pH=2\]You need to login to perform this action.
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