Question

Relative solubilities of salts in liquid ammonia can differ significantly from those in water. Thus, silver bromide is soluble in ammonia, but barium bromide is not (the reverse of the situation in water).

(a) Write a balanced equation for the reaction of an ammonia solution of barium nitrate with an ammonia solution of silver bromide. Silver nitrate is soluble in liquid ammonia.

(b) What volume of a $0.50\text{M}$ solution of silver bromide will react completely with $0.215L$of a $0.076\text{M}$solution of barium nitrate in ammonia?

(c) What mass of barium bromide will precipitate from the reaction in part (b)?

Short answer

(a) The balanced equation is:

$\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{in}\right.\text{liq}\left.{\text{NH}}_{\text{3}}\right)\text{+ 2AgBr}\left(\text{in}\right.\text{liq}\left.{\text{NH}}_{\text{3}}\right)\to {\text{2AgNO}}_{\text{3}}\left(\text{in}\right.\text{liq}\left.{\text{NH}}_{\text{3}}\right){\text{+ BaBr}}_{\text{2}}\text{(s)}$

(b) 0.065 L.

(c) $4.8g$.

Step-by-step solution

Concept of relative solubility

Solubility is the solute’s relative ability to be dissolved into a solvent. Relative solubility is the “maximal amount” of the substance in grams that can be dissolved in 100 grams of the solvent.

$0.50M$ solution of silver bromide will react completely with $0.215L$ of a $00076M$ solution of barium nitrate in ammonia.

Balanced equation for the given expression

(a)

A balanced equation for the reaction of an ammonia solution of barium nitrate with an ammonia solution of silver bromide is given below:

$\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{in}\right.\text{liq}\left.{\text{NH}}_{\text{3}}\right)\text{+ 2AgBr}\left(\text{in}\right.\text{liq}\left.{\text{NH}}_{\text{3}}\right)\to {\text{2AgNO}}_{\text{3}}\left(\text{in}\right.\text{liq}\left.{\text{NH}}_{\text{3}}\right){\text{+ BaBr}}_{\text{2}}\text{(s)}$

Calculate the volume of AgBr react with BaNO3

(b) The moles of barium nitrate is calculated as,

According to the balanced equation;

1 mole of $\text{Ba}{\left({\text{NO}}_3\right)}_2=2$ moles of $\mathrm{AgBr}$

Therefore,

Volume of AgBr is evaluated as:

Calculate the mass of  precipitation

(c)

According to the equation;

1 mole of $\text{Ba}{\left({\text{NO}}_3\right)}_2=1$mole of $\text{BaBr}{r}_2$

Therefore,

0.01634 moles of$\text{Ba}{\left({\text{NO}}_3\right)}_2=0.01634$moles of${\text{BaBr}}_2$

role="math" localid="1663412365751" $$\begin{gathered} {\text{Molar mass of BaBr}}_2=297.14\text{g}/\text{mol} \\ {\text{Mass of BaBr}}_2=\text{Moles}\times \text{Molar mass} \\ =0.01634\text{mol}\times 297.14\text{g}/\text{mol} \\ =\mathbf{4}.\mathbf{8}\mathbf{g} \end{gathered}$$