A 2.0 mL sample of $N a O H$ $NaOH$ solution is exactly neutralized by 4.0 mL of 3.0 M $H C l$ $HCl$ solution. What is the concentration of the $N a O H$ $NaOH$ solution?

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A 2.0 mL sample of $N a O H$ $NaOH$ solution is exactly neutralized by 4.0 mL of 3.0 M $H C l$ $HCl$ solution. What is the concentration of the $N a O H$ $NaOH$ solution?

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Answer 1 · 2016-06-03T06:04:14

$[N a O H]$ $[NaOH]$ $=$ $=$ $6.0 \cdot m o l \cdot L^{- 1}$ $6.0*mol*L^-1$

Explanation:

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

Clearly a 1:1 stoichiometry pertains. We know the concentration of the hydrochloric acid, and we use this to calculate (i) the number of moles of hydrochloric acid, and (ii) the concentration of the caustic soda.

$Moles of hydrochloric acid$ $"Moles of hydrochloric acid"$ $=$ $=$ $4.0 \times 10^{- 3} L \times 3.0 \cdot m o l \cdot L^{- 1}$ $4.0xx10^-3Lxx3.0*mol*L^-1$ $=$ $=$ $12 \times 10^{- 3} \cdot m o l$ $12xx10^-3*mol$

$Concentration of sodium hydroxide$ $"Concentration of sodium hydroxide"$ $=$ $=$ $\frac{Moles of NaOH}{Initial volume of solution}$ $"Moles of NaOH"/"Initial volume of solution"$ $=$ $=$ $\frac{12 \times 10^{- 3} \cdot m o l}{2.0 \times 10^{- 3} \cdot L}$ $(12xx10^-3*mol)/(2.0xx10^-3*L)$

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A 2.0 mL sample of $N a O H$ $NaOH$ solution is exactly neutralized by 4.0 mL of 3.0 M $H C l$ $HCl$ solution. What is the concentration of the $N a O H$ $NaOH$ solution?

1 Answer

Explanation:

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A 2.0 mL sample of NaOHNaOH solution is exactly neutralized by 4.0 mL of 3.0 M HClHCl solution. What is the concentration of the NaOHNaOH solution?

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A 2.0 mL sample of $N a O H$ $NaOH$ solution is exactly neutralized by 4.0 mL of 3.0 M $H C l$ $HCl$ solution. What is the concentration of the $N a O H$ $NaOH$ solution?