An initial 25.00 ml solution of nitric acid (hno3) was diluted to 250.00 ml, and15.00 ml of this diluted solution was titrated with a 0.2244 m solution of sodium hydroxide. if 21.33 ml of the sodium hydroxide solution were required to neutralize the acid, what is the nitric acid concentration of the initial 25.00 ml solution?

HNO3 concentration in the initial 25.00 mL solution is 3.19 M

Explanation:

We are working backward:

First we will determine the [HNO3] in 15 mL that reacted with Sodium hydroxide (NaOH).

mole of nitric acid = mole of NaOH

[HNO3] * 15 mL = 0.2244 M * 21.33 mL

[HNO3] = 0.319 M

The above is the [HNO3] in 250 mL solution that was transferred as 15 mL to react with NaOH.

Second, we can determine the [HNO3] in 25 mL because the 250 mL solution was made from the 25 mL.

[HNO3] * 25 mL = 0.319 M * 250 mL

[HNO3] = 3.19 M which is the concentration in the initial 25 mL solution.  

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