What volume of $N a O H (a q)$ $NaOH(aq)$ at $0.75 \cdot m o l \cdot L^{- 1}$ $0.75molL^-1$ concentration is required to neutralize a $275 \cdot m L$ $275mL$ volume of $H_{2} S O_{4} (a q)$ $H_2SO_4(aq)$ whose concentration is $0.50 \cdot m o l \cdot L^{- 1}$ $0.50mol*L^-1$ ?

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What volume of $N a O H (a q)$ $NaOH(aq)$ at $0.75 \cdot m o l \cdot L^{- 1}$ $0.75molL^-1$ concentration is required to neutralize a $275 \cdot m L$ $275mL$ volume of $H_{2} S O_{4} (a q)$ $H_2SO_4(aq)$ whose concentration is $0.50 \cdot m o l \cdot L^{- 1}$ $0.50mol*L^-1$ ?

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Answer 1 · 2016-05-19T06:14:28

Approx. $400 \cdot m L$ $400*mL$

Explanation:

$2 N a O H (a q) + H_{2} S O_{4} (a q) \to N a_{2} S O_{4} (a q) + 2 H_{2} O (l)$ $2NaOH(aq) + H_2SO_4(aq) rarr Na_2SO_4(aq) + 2H_2O(l)$

There is a $2 : 1$ $2:1$ stoichiometry between between caustic soda and sulfuric acid.

$Moles of sulfuric acid$ $"Moles of sulfuric acid"$ $=$ $=$ $0.50 \cdot m o l \cdot L^{- 1} \times 0.275 \cdot L$ $0.50*mol*L^-1xx0.275*L$ $=$ $=$ $0.138 \cdot m o l$ $0.138*mol$ .

Thus $0.275 \cdot m o l$ $0.275*mol$ $sodium hydroxide$ $"sodium hydroxide"$ are required.

$(0.275*cancel(mol))/(0.75*cancel(mol*L^-1))xx10^3*mL*cancelL^-1$ $=$ $??*mL$

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What volume of $N a O H (a q)$ $NaOH(aq)$ at $0.75 \cdot m o l \cdot L^{- 1}$ $0.75molL^-1$ concentration is required to neutralize a $275 \cdot m L$ $275mL$ volume of $H_{2} S O_{4} (a q)$ $H_2SO_4(aq)$ whose concentration is $0.50 \cdot m o l \cdot L^{- 1}$ $0.50mol*L^-1$ ?

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What volume of $N a O H (a q)$ $NaOH(aq)$ at $0.75 \cdot m o l \cdot L^{- 1}$ $0.75molL^-1$ concentration is required to neutralize a $275 \cdot m L$ $275mL$ volume of $H_{2} S O_{4} (a q)$ $H_2SO_4(aq)$ whose concentration is $0.50 \cdot m o l \cdot L^{- 1}$ $0.50mol*L^-1$ ?