A buffer solution contains 0.329 M NaHCO3 and 0.246 M Na2CO3. Determine the pH change when 0.072 mol KOH is added to 1.00 L of the buffer. pH after ad

The pH change is 0.22.

Explanation:

The pKa of the buffer NaHCO3/Na2CO3 is 10.329. To find the pH of this buffer we can use H-H equation:

pH = pKa + log [Na2CO3] / [NaHCO3]

Where pH is the pH of the buffer, pKa is 10.329 and [] could be taken as the moles of each specie.

As the initial volume of the buffer is 1.00L, the initial moles are:

[Na2CO3] = 0.246mol

[NaHCO3] = 0.329mol

Initial pH is:

pH = 10.329 + log [0.246mol] / [0.329mol]

pH = 10.20

The KOH reacts with NaHCO3 as follows:

NaHCO3 + KOH → NaKCO3 + H2O

Where NaKCO3≅ Na2CO3. That means the moles added of KOH are the increasing in moles of KOH and the moles of NaHCO3 that decreases.

Thus, moles after the addition of KOH are:

[Na2CO3] = 0.246mol + 0.072mol = 0.318mol

[NaHCO3] = 0.329mol - 0.072mol = 0.257mol

And pH is:

pH = 10.329 + log [0.318mol] / [0.257mol]

pH = 10.42

The pH change is:

10.42 - 10.20 =

0.22

answer

the ions in solute supperate from eachouthere