What volume of 0.500 M $N a O H$ $NaOH$ is needed to neutralize 45.0 mL of 0.400 M $H C l$ $HCl$ ?

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What volume of 0.500 M $N a O H$ $NaOH$ is needed to neutralize 45.0 mL of 0.400 M $H C l$ $HCl$ ?

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Answer 1 · 2016-05-03T15:47:23

$N a O H (a q) + H C l (a q) \to N a C l (a q) + H_{2} O (l)$ $NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)$

Explanation:

Clearly there is a 1:1 equivalence, and as a first step we calculate the number of moles of hydrochloric acid:

$45.0 \times 10^{- 3} \cdot L \times 0.400 \cdot m o l \cdot L^{- 1}$ $45.0xx10^-3*Lxx0.400*mol*L^-1$ $=$ $=$ $1.80 \times 10^{- 2} \cdot m o l hydrochloric acid$ $1.80xx10^-2*mol" hydrochloric acid"$ .

We find an equivalent molar quantity of sodium hydroxide:

$\frac{1.80 \times 10^{- 2} \cdot m o l}{0.500 \cdot m o l \cdot L^{- 1}} \times 10^{3} \cdot m L \cdot L^{- 1}$ $(1.80xx10^-2*mol)/(0.500*mol*L^-1)xx10^3*mL*L^-1$ $=$ $=$ $36.0 \cdot m L$ $36.0*mL$ .

This volume is reasonable, in that it is LESS than the volume of the less concentrated acid.

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What volume of 0.500 M $N a O H$ $NaOH$ is needed to neutralize 45.0 mL of 0.400 M $H C l$ $HCl$ ?

1 Answer

Explanation:

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What volume of 0.500 M NaOHNaOH is needed to neutralize 45.0 mL of 0.400 M HClHCl?

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Explanation:

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What volume of 0.500 M $N a O H$ $NaOH$ is needed to neutralize 45.0 mL of 0.400 M $H C l$ $HCl$ ?