50 mL of 0.1 M CH 3 COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be × 10-2 . (Nearest integer) (Given: .pK a ( CH 3 COOH )=4.76) log 2=0.30 log 3=0.48 log 5=0.69 log 7=0.84 log 11=1.04

Question

50 mL of 0.1 M CH 3 COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be × 10-2 . (Nearest integer) (Given: .pK a ( CH 3 COOH )=4.76) log 2=0.30 log 3=0.48 log 5=0.69 log 7=0.84 log 11=1.04 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Moles of CH 3 COOH =5 m mole moles of NaOH =2.5 m mole <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/ccebe3c7c9b6f0f03868bb20f25ae1b9-.png" /> so buffer is formed pH = pKa + log ((2.5 / 75/2.5 / 75))= pKa pH =4.76 =476 × 10-2

Q. 50mL of 0.1MCH3​COOH is being titrated against 0.1MNaOH. When 25mL of NaOH has been added, the pH of the solution will be _________×10−2. (Nearest integer) (Given : pKa​(CH3​COOH)=4.76) log2=0.30 log3=0.48 log5=0.69 log7=0.84 log11=1.04

Moles of CH3​COOH=5m mole moles of NaOH=2.5m mole so buffer is formed pH=pKa+log(2.5/752.5/75​)=pKa pH=4.76 =476×10−2

Moles of $C H_{3} COO H = 5 m$ mole moles of $N a O H = 2.5 m$ mole

so buffer is formed
$p H = p K a + lo g (\frac{2.5/75}{2.5/75}) = p K a$
$p H = 4.76$
$= 476 \times 1 0^{- 2}$