Question

How many grams of sodium formate, NaCHO2, would have to be dissolved in 3.0 L of 0.12 M formic acid (pKa 3.74) to make the solution a buffer for pH 5.30?

Answer 1

`Mass_(sodium\ formate)= 889.57\ g`

Step-by-step explanation:

Considering the Henderson- Hasselbalch equation for the calculation of the pH of the acidic buffer solution as:

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

Given that:-

[Acid] = 0.12 M

Volume = 3.0 L

pKa = 3.74

pH = 5.30

So,

`5.30=3.74+log([sodium\ formate])/(0.12)`

Solving, we get that:-

[Sodium formate] = 4.36 M

Considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

So,

`Moles =Molarity * {Volume\ of\ the\ solution}`

So, Moles of sodium formate = 4.36*3.0 moles = 13.08 moles

Molar mass of sodium formate = 68.01 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`13.08\ mole= (Mass)/(68.01\ g/mol)`

`Mass_(sodium\ formate)= 889.57\ g`