Calculate the pH of a buffer made

[kk727]

Homework Statement

Calculate the pH of a buffer made with 10.0 g of K2HPO4 and 5.0 g of KH2PO4 in 1.0L of water.

Homework Equations

-log Ka + log (base/acid) = pH

The Attempt at a Solution

So...mathwise, I know how to do this problem. My only question is, aren't K2HPO4 and KH2PO4 both salts? How would they break up? I don't know which would be the acid or the base, and I couldn't find a Ka/Kb value to help me.

Basically I just can't figure out how they break up and what their conjugates are... :p

 

[Borek]

Ignore cation. Write stepwise dissociation reactions for multiprotic acid.

 

[kk727]

So my teacher said that the KH2PO4 would be considered the acid, and that the K2HPO4 would form a strong conjugate base. I converted from grams to mols, and then to molarity. Using the given Ka value of 6.2 x 10^-8, I just plugged everything into the Henderson-Hasselbalch equation and got a pH of 7.4. Does this sound right? :/

 

[Borek]

Yes.

Do you understand why H₂PO₄^- is an acid?

 

[DeeNos]

It is important to first determine the nature of the salts K2HPO4 and KH2PO4 in order to understand how they will behave in solution. K2HPO4 is a salt of a weak acid (H3PO4) and a strong base (KOH), while KH2PO4 is a salt of a strong acid (H3PO4) and a weak base (KOH). This means that K2HPO4 will act as a weak base in solution, while KH2PO4 will act as a weak acid.

To find the pH of the buffer, we can use the Henderson-Hasselbalch equation:

pH = pKa + log ([base]/[acid])

The pKa for phosphoric acid (H3PO4) is 2.12, so we can plug this value in along with the concentrations of K2HPO4 and KH2PO4 in the buffer (calculated from the given amounts and volume of water). This will give us:

pH = 2.12 + log ([K2HPO4]/[KH2PO4])

To find the concentrations of K2HPO4 and KH2PO4, we can use the fact that in a 1:1 molar ratio, they will both contribute equally to the buffer solution. This means that the total concentration of the buffer (10.0 g + 5.0 g = 15.0 g) will be equally divided between the two salts, giving us a concentration of 7.5 g/L for each salt. Converting this to moles and dividing by the volume of the buffer (1.0 L), we get a concentration of 0.075 mol/L for each salt.

Plugging this into the Henderson-Hasselbalch equation, we get:

pH = 2.12 + log (0.075/0.075) = 2.12 + log (1) = 2.12

Therefore, the pH of the buffer solution will be 2.12. This means that the buffer is slightly acidic, which makes sense since it is made up of a weak base (K2HPO4) and a weak acid (KH2PO4).