Question

at 298 k solubility of silver chloride is 1.37 ×10-5 m .calculate its solubility of agcl in 1 m agno3 solution

Answer 1

+(aq) in a strong solution of ammonia. Calculate the molar solubility of AgCl in 3.0 M NH3. Ksp of AgCl = 1.8×10–10.

Answer 2

Answer:

The solubility of AgCl in 1 m $A g N o_{3}$ is $\bold{1.2 \times 10^{-9}}$

Explanation:

Given:

The solubility of silver chloride is $1.37 \times 10^{-5}$ m at the temperature of 298 k.

To find:

The solubility of AgCl in 1m $A g N o_{3}$

Let us consider the solubility of AgCl is s.  

So solubility of each ion will be:

Step 1:

i. NaCl gives sodium ion and chloride ion.  

ii. Here chloride ion is common so that the solubility of AgCl decreases, because of the presence of the common ion.

Step 2:

i. Molarity of NaCl is 0.1 M.

ii.    The total solubility of chloride ion in AgCl is (0.1+s).  

Step 3:

$\mathrm{Ksp}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$

Let us substitute the known values,  

We get,

$1.2 \times 10^{-10}=s \times(0.1+s)$

When the ksp value is low then the value of s can be neglected.

Therefore,

$1.2 \times 10^{-10}=s \times(0.1+s)$

Step 4:

To find the value of s

$\begin{array}{l}{\mathrm{S}=1.2 \times 10^{-10} / 0.1} \\ {=1.2 \times 10^{-9}}\end{array}$