Question

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

Answer 1

Molar solubility of AgBr in a solution that contains 0.200 m NaBr is $\boxed{{\text{3}}{\text{.8}}\times{{10}^{-12}}{\text{M}}}$.

Further explanation:

Molar solubility is the concentration of pure substance dissolved in saturated solution. It is generally defined as ratio of number of moles of substance dissolved in one litre of solution until the solution becomes saturated. The S.I unit of molar solubility is mol/L.

The equilibrium constant between the compound and its ion, when dissolved in solution, is known as solubility product constant. It is denoted by ${{\text{K}}_{{\text{sp}}}}$. The solubility product constant is used to calculate the product of concentration of ions at equilibrium. Higher the solubility product constant more will be the solubility of the compound.

The general reaction is as follows:

${\text{AB}}\left({{\text{aq}}}\right)\to{{\text{A}}^+}\left({{\text{aq}}}\right)+{{\text{B}}^-}\left({{\text{aq}}}\right)$

The expression to calculate solubility product for the general reaction is as follows:

${{\text{K}}_{{\text{sp}}}}=\left[{{{\text{A}}^+}}\right]\left[{{{\text{B}}^-}}\right]$

Here,

$\left[{{{\text{A}}^+}}\right]$. is the concentration of ${{\text{A}}^+}$ions.

$\left[{{{\text{B}}^-}}\right]$is the concentration of ${{\text{B}}^-}$ions.

Given,

The concentration of NaBr is 0.200 M.

The reaction of dissociation of NaBr in solution is written as follows:

${\text{NaBr}}\left({{\text{aq}}}\right)\to{\text{N}}{{\text{a}}^+}\left({{\text{aq}}}\right)+{\text{B}}{{\text{r}}^-}\left({{\text{aq}}}\right)$

0.200 M NaBr dissociates to give 0.200 M ${\text{N}}{{\text{a}}^+}$and 0.200 M ${\text{B}}{{\text{r}}^-}$.

The reaction of dissociation of AgBr in solution is written as follows:

${\text{AgBr}}\left({\text{s}}\right)\rightleftharpoons{\text{A}}{{\text{g}}^+}\left({{\text{aq}}}\right)+{\text{B}}{{\text{r}}^-}\left({{\text{aq}}}\right)$

Consider the molar solubility of AgBr be S. Therefore, after dissociation the concentration of ${\text{A}}{{\text{g}}^+}$and ${\text{B}}{{\text{r}}^-}$is S.

Both NaBr and AgBr in a solution contain ${\text{B}}{{\text{r}}^-}$ ions.

So, the total concentration of Br ion in solution is calculated as follows:

$\begin{aligned}{\text{Total concentration of B}}{{\text{r}}^-}&={\left[{{\text{B}}{{\text{r}}^-}}\right]_{{\text{NaBr}}}}+{\left[{{\text{B}}{{\text{r}}^-}}\right]_{{\text{AgBr}}}}\\&={\text{0}}{\text{.200}}+{\text{S}}\\\end{aligned}$

The formula to calculate the molar solubility of AgBr is as follows:

${{\text{K}}_{{\text{sp}}}}=\left[{{\text{A}}{{\text{g}}^{\text{+}}}}\right]\left[{{\text{B}}{{\text{r}}^-}}\right]$                       …… (1)

Substitute S for$\left[{{\text{A}}{{\text{g}}^{\text{+}}}}\right]$, 0.200 + S for$\left[{{\text{B}}{{\text{r}}^-}}\right]$ and${\text{7}}{\text{.7}}\times{10^{-13}}$for ${{\text{K}}_{{\text{sp}}}}$in equation (1).

$\begin{aligned}{\text{7}}{\text{.7}}\times{10^{-13}}=\left[{\text{S}}\right]\left[{{\text{0}}{\text{.200}}+{\text{S}}}\right]\hfill\\7.7\times{10^{-13}}={{\text{S}}^2}+0.{\text{200S}}\hfill\\{{\text{S}}^2}+0.{\text{200S}} -7.7\times{10^{-13}}=0\hfill\\\end{aligned}$

After solving the quadratic equation the value of S obtained is as follows:

$\left[{\text{S}}\right]={\text{3}}{\text{.8}}\times{10^{-12}}{\text{M}}$

The molar solubility of AgBr in solution is ${\mathbf{3}}{\mathbf{.8\times1}}{{\mathbf{0}}^{{\mathbf{-12}}}}{\mathbf{M}}$.

Learn more:

1. Sort the solubility of gas will increased or decreased: brainly.com/question/2802008.

2. What is the pressure of gas?: brainly.com/question/6340739.

Answer details:

Grade: School school

Subject: Chemistry

Chapter: Chemical equilibrium

Keywords: Molar solubility, equilibrium, litre, mole, amount, substance, product, concentration, initial; dissolved, saturated solution, stoichiometry, S, NaBr and AgBr.

Answer 2

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.