Question

Find pH of 0.003 M of Ch3COOH whose Ka =1.8×10^-5​

Answer 1

hi buddy

here is your answer

• Ka of acetic acid is 1.8 x 10^-5Ka = [H3O+][CH3COO-]/[CH3COOH]1.8 x 10^-5 = x.x/(0.003)On solving, we get, x = 2.32 x 10^-4[H+] = x = 2.32 x 10^-4pH = -log[H+] = -log (2.32 x 10^-4) = 3.63

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Answer 2

Answer:

• pH = 3.64

Steps:

pH of a particular substance is calculated by the formula:

• pH = - log [ H⁺ ]

To Calculate [ H⁺ ], we use the formula:

→ [ H⁺ ] = √ (Ka × C)

Where, Ka is the ionisation constant and C is known as the Concentration of  the Acid / Base given.

According to the question,

Ka = 1.8 × 10⁻⁵ & C = 0.003 M

⇒ [ H⁺ ] = √ ( 1.8 × 10⁻⁵ × 0.003 )

⇒ [ H⁺ ] = √ ( 5.4 × 10⁻⁸ )

⇒ [ H⁺ ] = 2.32 × 10⁻⁴

Calculating the pH we get:

⇒ pH = - log [ 2.32 × 10⁻⁴ ]

⇒ pH = - log ( 2.32 ) - ( -4 ) log ( 10 )

⇒ pH = -0.36 + 4 ( 1 )

⇒ pH = 3.64

Hence the pH of 0.003 M of CH₃COOH is 3.64