SATHEE: Related Problems with Solution

Related Problems with Solution

Problem 1 : Calculate the pH of a 0.02 M solution of NH₄CN. Given K_b for NH₄CN is $(2.0 \times 10^{- 5}) M .$

Solution :

We can use the K_b expression for the ionization of NH₄CN: $[K_{b} = \frac{[N H_{4}^{+}] [C N^{-}]}{[N H_{4} C N]}]$

Since we have K_b and the initial concentration of NH₄CN., we can set up an ICE (initial, change, equilibrium) table to calculate the concentration of (OH^-) ions and then find (pOH):


  NH4CN   =>   NH4+   +   CN-
---------------------------------
Initial    0.02 M       0 M     0 M
Change     -x           x       x
Equilibrium 0.02 - x     x       x

From the K_b expression, we have: $[2.0 \times 10^{- 5} = \frac{x \cdot x}{0.02 - x}]$

Since (x) is small compared to 0.02, we can approximate (0.02 - x) as 0.02: $[2.0 \times 10^{- 5} = \frac{x^{2}}{0.02}]$

Now, solve for (x): $[x^{2} = 2.0 \times 10^{- 5} \cdot 0.02]$ $[x^{2} = 4.0 \times 10^{- 7}]$ $[x = \sqrt{4.0 \times 10^{- 7}}]$ $[x = 2.0 \times 10^{- 4}, M]$

Now that we have the OH^- ion concentration, we can calculate (pOH): $[p O H = - \log (2.0 \times 10^{- 4})]$

Calculate (pOH): $[p O H \approx 3.70]$

Finally, we can find the pH using the relation: (pH + pOH = 14): $[p H = 14 - p O H]$ $[p H \approx 14 - 3.70]$ $[p H \approx 10.30]$

So, the pH of the solution is approximately 10.30.

Related Problems with Solution

Problem 1 : Calculate the pH of a 0.02 M solution of NH4CN. Given Kb for NH4CN is (2.0×10−5)M.

Solution :

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

Menu

Problem 1 : Calculate the pH of a 0.02 M solution of NH₄CN. Given K_b for NH₄CN is $(2.0 \times 10^{- 5}) M .$