SATHEE: Related Problems with Solution

Related Problems with Solution

Problem 2 : Calculate the pH of a 0.1 M solution of (NaF). Given K_a for (HF) is $(6.8 \times 10^{- 4}) .$

Solution :

The ionization of (NaF) in water is as follows: $[N a F ⇌ N a^{+} + F^{-}]$

We can use the K_a expression for the ionization of (HF): $[K_{a} = \frac{[H^{+}] [F^{-}]}{[H F]}]$

Since (NaF) is a salt, it dissociates completely into its ions. So, the concentration of F^- ions from (NaF) is 0.1 M.

Now, we can set up an ICE table for the ionization of (HF):

  HF   =>   H+   +   F-
----------------------------
Initial    0 M      0 M     0 M
Change      -x       x       x
Equilibrium -x       x       x

From the K_a expression, we have: $[6.8 \times 10^{- 4} = \frac{x \cdot x}{0.1 - x}]$

Since (x) is small compared to 0.1, we can approximate (0.1 - x) as 0.1: $[6.8 \times 10^{- 4} = \frac{x^{2}}{0.1}]$

Now, solve for (x): $[x^{2} = 6.8 \times 10^{- 4} \cdot 0.1]$ $[x^{2} = 6.8 \times 10^{- 5}]$ $[x = \sqrt{6.8 \times 10^{- 5}}]$ $[x \approx 8.26 \times 10^{- 3}, M]$

Now, calculate the pH using the H⁺ ion concentration: $[p H = - \log (8.26 \times 10^{- 3})]$

Calculate (pH): $[p H \approx 2.08]$

So, the pH of the solution is approximately 2.08.

Related Problems with Solution

Problem 2 : Calculate the pH of a 0.1 M solution of (NaF). Given Ka for (HF) is (6.8×10−4).

Solution :

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

Menu

Problem 2 : Calculate the pH of a 0.1 M solution of (NaF). Given K_a for (HF) is $(6.8 \times 10^{- 4}) .$