What is pH of a buffer made by combining 45.0mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate

3.11

Explanation:

Any buffer system can be described with the reaction:

Where is the acid and is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:

With all this in mind, we can write the reaction for our buffer system:

-) Nitrous acid:

-) Sodium nitrate:

In this case, the acid is with a concentration of 0.150 M and a volume of 45.0 mL (0.045 L). The base is with a concentration of 0.175 M and a volume of 20.0 mL (0.020 L).

We can calculate the moles of each compound is we take into account the molarity equation (). So:

-) moles of :

-) moles of :

The total volume would be:

0.020 L + 0.045 L = 0.065 L

With this in mind, we can calculate the molarity of each compound:

-) Concentration of

-) Concentration of

The pKa reported is 3.39, therefore we can plug the values into the Henderson-Hasselbach equation:

The final pH value would be 3.11

I hope it helps!

the awnser is group 12