Determine the molar solubility of agbr in a solution containing 0.200 m nabr. ksp (agbr) = 7.7 × 10-13. determine the molar solubility of agbr in a solution containing 0.200 m nabr. ksp (agbr) = 7.7 × 10-13. 5.8 × 10-5 m 0.200 m 1.54 × 10-13 m 8.8 × 10-7 m 3.8 × 10-12 m

NaBr gives fully seperated ions as Na⁺(aq) and Br⁻(aq) while AgBr gives partially dissociated ions as Ag⁺(aq) and Br⁻(aq).
                 
              NaBr(aq) →Na⁺(aq) + Br⁻(aq)
              0.200                        0.200
             
                    AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)
Initial                               -               -
Change         -X            +X            +X
Equilibrium                     X              X

Ksp              =       [Ag⁺(aq)] [ Br⁻(aq)]
7.7 × 10⁻¹³   =        X * (0.200 + X)
7.7 × 10⁻¹³   =        X² + 0.200X
       0           =        X² + 0.200X-7.7 × 10⁻¹³
X1 = 3.8 x 10⁻¹²
X2 = -0.2

X = molar solubility of AgBr.
hence, X cannot be a negative value.

molar solubility of AgBr is 3.8 x 10⁻¹² M.

Density= mass/volume so 5.1 divided by 4.7 is  5.1 / 4.7 =1.08510638298round to the correct number of sig figs