Given aqueous solution of 0.1 m acetic acid of pka 4.72 and 0.1 m od sodium acettate, what volumes of actetic acid and sodium actetate should be mixed to make a 50 ml of a buffer soilution with ph 5.2?

12.44 mL of 0.1 M acetic acid solution

37.56 mL of 0.1 M sodium acetate solution

Explanation:

To solve this problem we can use Henderson-Hasselbach's equation:

Where [A⁻] and [HA] are the molar concentrations of sodium acetate and acetic acid, respectively.

First we use the equation to calculate the molar concentration ratio in the buffer solution:

We can rewrite the last part, using the definition of molar concentrations (C=n/V):

From the problem we also know that:

We rewirte the equation once again, keeping in mind that the moles of each substance come from the original 0.1 M solutions:

So now we have a system of two equations and two unknowns, we use algebra to solve them:

We replace that value in the other equation:

We use that value in the first equation:

Finally, we use the definition of molarity to calculate the volume of each solution:

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3