Answer:
0.288g KBT can be dissolved
Step-by-step explanation:
The dissolution of KBT produce the dissociation in K⁺ and BT⁻ ions as follows:
KBT(s) → K+(aq) + BT-(aq) Ksp = 3.8×10-4
Where Ksp is defined as:
Ksp = 3.8x10⁻⁴ = [K⁺][BT⁻]
Thus, we need to find the concentration of KBT that can be dissolved = [K⁺] = [BT⁻]:
3.8x10⁻⁴ = [K⁺][BT⁻]
3.8x10⁻⁴ = [KBT][KBT]
3.8x10⁻⁴ = [KBT]²
0.01949M = [KBT]
In 280mL = 0.280L, the moles of KBT that can be dissolved are:
0.01949mol / L * 0.280L =
5.458x10⁻³ moles of KBT. In grams (Using molar mass):
5.458x10⁻³ moles of KBT * (188.2g/mol) =
0.288g KBT can be dissolved