In this acid base reaction: $N a O H + H C l \to N a C l + H_{2} O$ $NaOH+HCl->NaCl+H_2O$ , if we measured out 10.0 mL of a 2.0 M solution of $H C l$ $HCl$ , how many grams of sodium hydroxide( $N a O H$ $NaOH$ ; MW=40 g/mol) would we need to neutralize the $H C l$ $HCl$ solution?

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In this acid base reaction: $N a O H + H C l \to N a C l + H_{2} O$ $NaOH+HCl->NaCl+H_2O$ , if we measured out 10.0 mL of a 2.0 M solution of $H C l$ $HCl$ , how many grams of sodium hydroxide( $N a O H$ $NaOH$ ; MW=40 g/mol) would we need to neutralize the $H C l$ $HCl$ solution?

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Answer 1 · 2016-09-27T06:58:03

$0.8 \cdot g$ $0.8*g$ of sodium hydroxide are required.

Explanation:

You have the stoichiometric equation. There is clearly 1:1 equivalence. Thus $moles of hydrochloric acid$ $"moles of hydrochloric acid"$ $=$ $=$ $moles of sodium hydroxide$ $"moles of sodium hydroxide"$ .

$Moles of hydrochloric acid$ $"Moles of hydrochloric acid"$ $=$ $=$ $10.0 \times 10^{- 3} L \times 2.0 \cdot m o l \cdot L^{- 1}$ $10.0xx10^-3Lxx2.0*mol*L^-1$ $=$ $=$ $0.020$ $0.020$ $m o l$ $mol$ $H C l$ $HCl$ .

$Mass of sodium hydroxide req'd$ $"Mass of sodium hydroxide req'd"$ $=$ $=$ $0.020 \cdot m o l \times 40.00 \cdot g \cdot m o l^{- 1}$ $0.020*molxx40.00*g*mol^-1$ $=$ $=$ $0.80 \cdot g$ $0.80*g$ .

It would be hard to measure such a quantity of sodium hydroxide as it is deliquescent, and once removed from the tin (or if the tin is poorly sealed), it would absorb moisture from the atmosphere.

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In this acid base reaction: $N a O H + H C l \to N a C l + H_{2} O$ $NaOH+HCl->NaCl+H_2O$ , if we measured out 10.0 mL of a 2.0 M solution of $H C l$ $HCl$ , how many grams of sodium hydroxide( $N a O H$ $NaOH$ ; MW=40 g/mol) would we need to neutralize the $H C l$ $HCl$ solution?

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Explanation:

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In this acid base reaction: NaOH+HCl→NaCl+H2ONaOH+HCl->NaCl+H_2O, if we measured out 10.0 mL of a 2.0 M solution of HClHCl, how many grams of sodium hydroxide(NaOHNaOH; MW=40 g/mol) would we need to neutralize the HClHCl solution?

1 Answer

Explanation:

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In this acid base reaction: $N a O H + H C l \to N a C l + H_{2} O$ $NaOH+HCl->NaCl+H_2O$ , if we measured out 10.0 mL of a 2.0 M solution of $H C l$ $HCl$ , how many grams of sodium hydroxide( $N a O H$ $NaOH$ ; MW=40 g/mol) would we need to neutralize the $H C l$ $HCl$ solution?