20 mL of M / 10 CH 3 COOH solution is titrated with M/10 NaOH solution. After addition of 16 mL solution of NaOH. What is the pH of the solution ( p Ka=4.47)

Question

20 mL of M / 10 CH 3 COOH solution is titrated with M/10 NaOH solution. After addition of 16 mL solution of NaOH. What is the pH of the solution ( p Ka=4.47) (A) 5.05 (B) 4.15 (C) 4.75 (D) 5.35

Tardigrade Admin · Accepted Answer

mEq of CH 3 COOH =20 × (1/10)=2 mEq of NaOH =16 × (1/10)=1.6 So 1.6 m Eq of NaOH reacts with 1.6 m Eq of CH 3 COOH to form 1.6 m Eq of salt CH 3 COONa and (2-1.6=0.4 mEq ) of WA( CH 3 COOH ) is left and an acidic buffer is formed. So to calculate pH, buffer equation is used pH = p Ka+ log (( text Salt / text Acid ))=4.74+ log ((1.6/0.4)) =4.74+2 log 2 =4.74+2 × 0.3010=5.35

Q. $20 m L$ of $M /10 C H_{3} COO H$ solution is titrated with $M /10 N a O H$ solution. After addition of $16 m L$ solution of $N a O H$ . What is the $p H$ of the solution $(p K_{a} = 4.47)$

$5.05$

$4.15$

$4.75$

$5.35$

Q. 20mL of M/10CH3​COOH solution is titrated with M/10NaOH solution. After addition of 16mL solution of NaOH. What is the pH of the solution (pKa​=4.47)

5.05

4.15

4.75

5.35

Q. $20 m L$ of $M /10 C H_{3} COO H$ solution is titrated with $M /10 N a O H$ solution. After addition of $16 m L$ solution of $N a O H$ . What is the $p H$ of the solution $(p K_{a} = 4.47)$

$5.05$

$4.15$

$4.75$

$5.35$