Question

a volume of 75mL of 0.040M NaF is mixed with 25mL of 0.10M Sr(NO3)2. Calculate the concentration of the following ions in the final solution (Ksp for SrF2= 2.0x10^-10). [NO3-] [Na+] [Sr2+] [F-].

Answer 1

NaF----> Na+ +F- Sr(NO3)₂ ----> Sr+2 +2NO3-

Moles of NaF= moles of Na+ =moles of F- , moles of NaF= 0.04*75/1000=0.003

Moles of Sr+ = 0.1*25/1000=0.0025, moles of NO3- = 2*0.0025=0.005M

volume after mixing = 75+25=100ml =0.1L

Concentrations : [Na+] = [F-] =0.003/0.1 =0.03M, [Sr+] = 0.0025/0.1= 0.025 [NO3-] = 0.005/0.1= 0.05Mt

The ionic reaction is Sr+2 +2F-----> SrF2

1 moles of Sr+2 requires 2 moles of F-    Molar ratio as per reaction =1 :2 and given molar ratio = 0.0025:0.003 =1 : 1.2 So F- is limting reactant and it consumes 0.003/2=0.0015 moles of Sr+2 unreacted = 0.0025-0.0015= 0.001

Concentration of Sr+2 =0.001/0.1 =0.01M

at Equilibrium [F-] =2x and Sr+2 =0.01+x

KSp = [F-]2 [Sr+2] =4x²*(0.01+x)= 2*10^-10

negelcting x   gives x=7.07*10^-5   Sr+2 =0.01+7.07*10^-5 = 0.0100   [F-] = 2*7.07*10^-5 =14.14*10^-5M

[NO3-] =0.05M and Na+ =0.03M