Question

calculate the ph of 1.0 l of a buffer that is 0.100 m hno2 and 0.150 m nano2. what is the ph of the same buffer after the addition of 1.0 ml of 12 m h

Answer 1

The pH of the buffer solution is 3.38.

After the addition of 1.0 mL of 12 M HCl, the pH becomes 3.43.

Step-by-step explanation:

To calculate the pH of the buffer solution, you can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

In this case, the acid is HNO2 and the conjugate base is NO2-.

The pKa of HNO2 is 3.2.

So, plugging in the values:

pH = 3.2 + log(0.150/0.100)

= 3.2 + log(1.5)

= 3.2 + 0.18

= 3.38

After the addition of 1.0 mL of 12 M HCl, the new concentrations will be:

[HNO2] = 0.100 M - 1.0 mL x (12 M / 1000 mL)

= 0.100 M - 0.012 M

= 0.088 M

[NO2-] = 0.150 M

Using the Henderson-Hasselbalch equation again:

pH = pKa + log([NO2-]/[HNO2])

= 3.2 + log(0.150/0.088)

= 3.2 + log(1.70)

= 3.2 + 0.23

= 3.43