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Blizzard [7]
3 years ago
15

A volume of 25.0 mL of nitric acid, HNO3, is titrated with .12 M NaOH. To completely neutralize the acid 10.0 mL of NaOH must be

added. What is The molarity of nitric acid is
Chemistry
1 answer:
igor_vitrenko [27]3 years ago
5 0

Answer:

0.048M

Explanation:

Volume of acid = 25mL = 0.025L

Concentration of base NaOH = 0.12M

Volume of base NaOH = 10mL= 0.01L

Unknown:

concentration or molarity of the nitric acid = ?

Solution

We first write the balanced reaction equation:

           HNO₃ + NaOH → NaNO₃ + H₂O

In this problem, the focus is on the reacting acid and base which are on the left hand side.

Now, to find the molarity of the nitric acid that would be completely neutralized by the base, we adopt the mole concept.

First, we work from the known to the unknow. We can find the number of moles of the base given because its parameters are complete. The number of moles is given by the expression below:

        Number of moles = concentration x volume

To find the number of moles of the base, we put the parameters into the expression above:

Number of moles of NaOH = concentration of NaOH x volume of NaOH

                                              = 0.01 x0.12

                                              = 0.0012mol

From the number of moles of the base, we can find the unknown.

  In the reaction equation:

      1 mole of base neutralizes 1 mole of the acid

      0.0012mole of the base will neutralize 0.0012mole of the acid

Therefore, 0.0012mole of HNO₃ was neutralized.

Now we can calculate the molarity of the acid using the expression below:

     concentration of acid = \frac{ number of moles of acid}{volume of acid}

     concentration of acid = \frac{0.0012 }{0.025} = 0.048M

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Explanation:

We'll begin by calculating the number of mole in 42 g of baking soda (NaHCO₃). This can be obtained as follow:

Mass of NaHCO₃ = 42 g

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Mole of NaHCO₃ =?

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Mole of NaHCO₃ = 42/84

Mole of NaHCO₃ = 0.5 mole

Next, balanced equation for the reaction. This is given below:

NaHCO₃ + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O + CO₂

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of CO₂

Finally, we shall determine the number of mole of CO₂ produced by the reaction of 42 g (i.e 0.5 mole) of NaHCO₃. This can be obtained as follow:

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of CO₂.

Therefore, 0.5 mole of NaHCO₃ will also react to produce 0.5 mole of CO₂.

Thus, 0.5 mole of CO₂ was obtained from the reaction.

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