Perform the following calculations:

(a) Calculate \(\left[ {A{g^ + }} \right]\)in a saturated aqueous solution of\(AgBr\).

(b) What will \(\left[ {A{g^ + }} \right]\)be when enough \(KBr\)has been added to make \(\left[ {B{r^ - }} \right] = 0.050M\) ?

(c) What will \(\left[ {B{r^ - }} \right]\)be when enough \(AgN{O_3}\)has been added to make\(\left[ {A{g^ + }} \right] = 0.020M\)?

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

Expression for\(\left[ {{\rm{B}}{{\rm{r}}^ - }} \right]\)is\(({\rm{x}} + 0.05)\)- here\(0.05\)is coming from\({\rm{KBr}}\). However,\(x\)here is very, very small number, so we can neglect it to simplify the equation. If we substitute\({{\rm{K}}_{{\rm{sp}}}}\)for its value, we get:

\(5 \cdot {10^{ - 13}} = {\rm{x}} \cdot 0.05\)

\({\rm{x}} = \frac{{5 \cdot {{10}^{ - 13}}}}{{0.05}}\)

\({\rm{x}} = 1 \cdot {10^{ - 11}}\)

\(\left[ {{\rm{A}}{{\rm{g}}^ + }} \right] = 1 \cdot {10^{ - 11}}\)

Dissociation of

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

Expression for\(\left[ {{\rm{A}}{{\rm{g}}^ + }} \right]\)is\(({\rm{x}} + 0.02)\)- here\(0.02\)is coming from\({\rm{AgN}}{{\rm{O}}_3}\). However,\({\rm{x}}\)here is very, very small number, so we can neglect it to simplify the equation. If we substitute\({{\rm{K}}_{{\rm{sp}}}}\)for its value, we get:

\(\begin{array}{l}5 \cdot {10^{ - 13}} = 0.02 \cdot {\rm{x}}\\{\rm{x}} = \frac{{5 \cdot {{10}^{ - 13}}}}{{0.02}}\\{\rm{x}} = 1 \cdot {10^{ - 11}}\end{array}\)

\(\left[ {{\rm{B}}{{\rm{r}}^ - }} \right] = 2.5 \cdot {10^{ - 11}}\)

Finally we get,

a) \(\left[ {{\rm{A}}{{\rm{g}}^ + }} \right] = 7.07 \cdot {10^{ - 7}}\)

b) \(\left[ {{\rm{A}}{{\rm{g}}^ + }} \right] = {1.10^{ - 11}}\)

c) \(\left[ {{\rm{B}}{{\rm{r}}^ - }} \right] = 2.5 \cdot {10^{ - 11}}\)