Question

На нейтрализацию 20,00 мл азотной кислоты израсходовали 17,45 мл 0,25 М раствора гидроксида натрия. Найдите титр и концентрацию азотной кислоты.​

Answer 1

Ответ:

Для HNO3 и NaOH Cm=Cn

СnV(HNO3)=CnV(NaOH)

Cm=(17.45*0.25)/20=0,218 моль/литр

T=CmM/1000 T=(0.218*63)/1000=0,013734 г/мл

Answer 2

Calculation of Nitric Acid Titration

To find the titration and concentration of nitric acid, we can use the given information that 17.45 mL of 0.25 M sodium hydroxide solution was used to neutralize 20.00 mL of nitric acid.

The balanced chemical equation for the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH) is:

HNO3 + NaOH → NaNO3 + H2O

From the balanced equation, we can see that the mole ratio of nitric acid to sodium hydroxide is 1:1.

Using the formula:

Moles = Molarity × Volume (in liters)

We can calculate the moles of sodium hydroxide used in the reaction:

Moles of NaOH = Molarity × Volume = 0.25 mol/L × (17.45 mL / 1000) L

Now, we can use the mole ratio to find the moles of nitric acid:

Moles of HNO3 = Moles of NaOH

Finally, we can calculate the concentration of the nitric acid:

Concentration of HNO3 = Moles of HNO3 / Volume of HNO3 (in liters)

Let's perform the calculations.

Calculation Steps

1. Moles of NaOH: - Moles of NaOH = 0.25 mol/L × (17.45 mL / 1000) L - Moles of NaOH = 0.25 mol/L × 0.01745 L - Moles of NaOH = 0.0043625 moles

2. Moles of HNO3: - Moles of HNO3 = Moles of NaOH - Moles of HNO3 = 0.0043625 moles

3. Concentration of HNO3: - Concentration of HNO3 = Moles of HNO3 / Volume of HNO3 - Concentration of HNO3 = 0.0043625 moles / (20.00 mL / 1000) L - Concentration of HNO3 = 0.0043625 moles / 0.02000 L - Concentration of HNO3 = 0.218125 mol/L

Results

- Moles of NaOH: 0.0043625 moles - Moles of HNO3: 0.0043625 moles - Concentration of HNO3: 0.218125 mol/L

Therefore, the titration of nitric acid is 0.218125 mol/L, and the concentration of nitric acid is 0.218125 mol/L.