What reaction occurs when 25.0 mL of 0.600 M NaOH are mixed with 15.0 mL of 0.400 M HNO3? What is the balanced equation and the net ionic equation?

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What reaction occurs when 25.0 mL of 0.600 M NaOH are mixed with 15.0 mL of 0.400 M HNO3? What is the balanced equation and the net ionic equation?

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Answer 1 · 2016-03-08T17:59:35

The net ionic equation is simply:

$H_{3} O^{+} +^{-} O H \to 2 H_{2} O (l)$ $H_3O^+ + ""^(-)OH rarr 2H_2O(l)$

Explanation:

The sodium ions and the nitrate ions are simply along for the ride.

Of course, it is clear that there is an excess of base.

Moles of $N a O H$ $NaOH$ $=$ $=$ $25.0 \times 10^{- 3} \cdot L \times 0.600 \cdot m o l \cdot L^{- 1}$ $25.0xx10^-3*Lxx0.600*mol*L^-1$ $=$ $=$ $? ? m o l$ $??mol$

Moles of $H N O_{3}$ $HNO_3$ $=$ $=$ $15.0 \times 10^{- 3} \cdot L \times 0.400 \cdot m o l \cdot L^{- 1}$ $15.0xx10^-3*Lxx0.400*mol*L^-1$ $=$ $=$ $? ? m o l$ $??mol$

The balanced equation is:

$N a O H (a q) + H N O_{3} (a q) \to N a N O_{3} (a q) + H_{2} O (l)$ $NaOH(aq) + HNO_3(aq) rarr NaNO_3(aq) + H_2O(l)$ .

I leave it to you to calculate the end concentration of sodium hydroxide. Remember, you have added the volumes so now the volume of the solution is $40.0$ $40.0$ $m L$ $mL$ .

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What reaction occurs when 25.0 mL of 0.600 M NaOH are mixed with 15.0 mL of 0.400 M HNO3? What is the balanced equation and the net ionic equation?

1 Answer

Explanation:

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