1 × 10-3 mole of HCl is added t a buffer solution made up of 0.01 M acetic and 0.01 M sodium acetate. The final pH of the buffer will be (given, pKa of acetic acid is 4.75 at 25°C)

Question

1 × 10-3 mole of HCl is added t a buffer solution made up of 0.01 M acetic and 0.01 M sodium acetate. The final pH of the buffer will be (given, pKa of acetic acid is 4.75 at 25°C) (A) 4.60 (B) 4.66 (C) 4.75 (D) 4.8

Tardigrade Admin · Accepted Answer

CH3COO- + H+ → CH3 COOH <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/f67c95ee1619e4505cb42b35fb42b31a-.png" /> PH =PKa textlog ( text(salt)/ text(acid)) = 4.75+ textlog (0.009/0.011) =4.66

Q. $1 \times 1 0^{- 3}$ mole of HCl is added t a buffer solution made up of 0.01 M acetic and 0.01 M sodium acetate. The final pH of the buffer will be (given, $p K_{a}$ of acetic acid is 4.75 at 25 $^{\circ}$ C)

4.60

4.66

4.75

4.8

$C H_{3} CO O^{-} + H^{+} \to C H_{3} COO H$

$P H = P K_{a} log \frac{(salt)}{(acid)} = 4.75 + log \frac{0.009}{0.011}$
$= 4.66$

Q. 1×10−3 mole of HCl is added t a buffer solution made up of 0.01 M acetic and 0.01 M sodium acetate. The final pH of the buffer will be (given, pKa​ of acetic acid is 4.75 at 25∘C)

4.60

4.66

4.75

4.8

CH3​COO−+H+→CH3​COOH PH=PKa​log(acid)(salt)​=4.75+log0.0110.009​ =4.66

Q. $1 \times 1 0^{- 3}$ mole of HCl is added t a buffer solution made up of 0.01 M acetic and 0.01 M sodium acetate. The final pH of the buffer will be (given, $p K_{a}$ of acetic acid is 4.75 at 25 $^{\circ}$ C)

$C H_{3} CO O^{-} + H^{+} \to C H_{3} COO H$

$P H = P K_{a} log \frac{(salt)}{(acid)} = 4.75 + log \frac{0.009}{0.011}$
$= 4.66$