Question

How many grams are KBr are needed to make a saturated solution in 100g of water at 30c

Answer 1

Explanation:

To calculate that you need to know the value of the Kps (constant of solubility) at 30°C of KBr in water.

When the product of the concentrations of both ions (Br- and K+) equals the Kps, the solution is saturated.

Given the disociation of the salt in water:

$KBr \longrightarrow K^+ + Br^-$

The concentration of both ions are equal (1:1 ratio) and the same as the concentration of KBr added.

The Kps:

$Kps=[K^+][Br^-]$

$[KBr]=(Kps)^{1/2}$

[KBr] are moles of the salt per litre of water

Assuming a density of water of 1 g/cm3, 100 g are 0.1 litre

To calculate the mass:

$m=[KBr]*M*0.1 litre$

where M is the molecular weight of the salt.