Ksp of SrF2 in water is 8*10^-10. calculate the solubility in water and 0.1M NaF solution Share with your friends Share 1 Akanksha C. answered this Hi,SrF2 s ⇔Sr+2 + 2F-Ksp = Sr+2 F-2Let the solubility of SrF2 in 0.1 M NaF be x mol L-1 NaF being a strong electrolyte ionises completely and supply 0.1 mol L-1 F- Sr+2 = x mol L-1 F- = 2x + 0.1mol L-1= 0.1mol L-1Ksp = x0.128 ×10-10 = x ×10-2x = 8 ×10-8 mol L-1Regards 10 View Full Answer
