Question

Calculate pH of buffer solution composed of 0.1M weak base BOH

and 0.2M of its salt BA. [Kb = 1.8  10-5] [03]

Write relation between solubility and solubility product for salt

Al(OH)3.​

Answer 1

Explanation:

I hope friend that this helps you

Answer 2

Given :

Concentration of base = 0.1 M

Concentration of salt = 0.2 M

To Find :

(a) pH of buffer solution

(b) Relation between solubility and solubility product for salt Al(OH)₃.​

Solution :

(a) Using Henderson's equation to determine the pOH of a buffer solution consisting of weak base and its salt -

       pOH  =  p$K_b$ +  ㏒ $\frac{[salt]}{[base]}$

⇒     pOH  =  -log (1.8 × 10⁻⁵) + ㏒$\frac{0.2}{0.1}$

⇒     pOH  =  5 - ㏒(1.8) + ㏒ 2

⇒     pOH  =  5 - 0.26 + 0.3

⇒     pOH  =  5 + 0.04

∴      pOH  =  5.04

We know,

    pH  +  pOH  =  14

⇒               pH   =  14 - 5.04

∴                pH   =  8.96

Therefore, the pH of the buffer solution is 8.96

(b)  Let the solubility be 'S'

For a compound say $A_xB_y$, the value of solubility product ($K_{sp}$) is equal to $x^x y^y S^x^+^y$

Thus for Al(OH)₃,

Al(OH)₃   ----------->   Al³⁺  +  3OH⁻

Here, x = 1 and y = 3

Solubility product ($K_{sp}$)  = (1)¹ × (3)³ × $S^1^+^3$

Solubility product ($K_{sp}$)  = 27S⁴