Question

A buffer is made by dissolving 11.6 g of sodium dihydrogen phosphate, NaH2PO4 (Mr = 119.97701128 g/mol), and 20.1 g of disodium hydrogen phosphate, Na2HPO4 (Mr = 141.95884056 g/mol), in a liter of solution. What is the pH of the buffer? (Assume Ka for this equilibrium is 6.2 ✕ 10−8, and that Kw is 1.0 ✕ 10−14.)

Answer 1

7.37 is the pH of the buffer.

Step-by-step explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

`pH=pK_a+\log(([salt])/([acid]))`

`pH=pK_a+\log(([Na_2HPO_4])/([NaH_2PO_4]))`

We are given:

`K_a` = dissociation constant =
`K_a=6.2* 10^(-8)`

Moles of
`Na_2HPO_4=(20.1 g)/(141.95884056 g/mol)=0.1416 mol`

Moles of
`NaH_2PO_4=(11.6 g)/(119.97701128 g/mol)=0.09668 mol`

`[concentration]=(moles)/(volume (L))`

`[NaH_2PO_4]=(0.09668 mol)/(1 L)=0.09668 M` (acid)

`[Na_2HPO_4]=(0.1416 mol)/(1 L)=0.1416 M` (salt)

pH = ?

Putting values in above equation, we get:

`pH=-\log[6.2* 10^(-8)]+\log((0.1416 M)/(0.09668 M))\\\\pH=7.37`

7.37 is the pH of the buffer.