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

Answer:

$Mass_{sodium\ formate}= 889.57\ g$

Explanation:

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

$pH=pK_a+log\frac{[salt]}{[acid]}$

Given that:-

[Acid] = 0.12 M

Volume = 3.0 L

pKa = 3.74

pH = 5.30

So,  

$5.30=3.74+log\frac{[sodium\ formate]}{0.12}$

Solving, we get that:-

[Sodium formate] = 4.36 M

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

So,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$13.08\ mole= \frac{Mass}{68.01\ g/mol}$

$Mass_{sodium\ formate}= 889.57\ g$