Ionisation constant of CH 3 COOH is 1.7 × 10-5 and concentration of H +ions is 3.4 × 10-4. Then, find out initial concentration of CH 3 COOH molecules.

Question

Ionisation constant of CH 3 COOH is 1.7 × 10-5 and concentration of H +ions is 3.4 × 10-4. Then, find out initial concentration of CH 3 COOH molecules. (A)  3.4 × 10-4  (B)  3.4 × 10-3  (C)  6.8 × 10-4  (D)  6.8 × 10-3

Tardigrade Admin · Accepted Answer

CH 3 COOH' leftharpoons CH 3 COO -+ H + Ka =([ CH 3 COO -][ H +]/[ CH 3 COOH ]) Given that, [ CH 3 COO -]=[ H +]=3.4 × 10-4 M Ka for CH 3 COOH =1.7 × 10-5 CH 3 COOH is weak acid, so in it [ CH 3 COOH ] is equal to initial concentration. Hence, 1.7 × 10-6 =((3.4 × 10-4)(3.4 × 10-4)/[ CH 3 COOH ]) [ CH 3 COOH ] =(3.4 × 10-4 × 3.4 × 10-4/1.7 × 10-5) =6.8 × 10-3 M

Q. Ionisation constant of $C H_{3} COO H$ is $1.7 \times 1 0^{- 5}$ and concentration of $H^{+}$ ions is $3.4 \times 1 0^{- 4}$ . Then, find out initial concentration of $C H_{3} COO H$ molecules.

$3.4 \times 1 0^{- 4}$

$3.4 \times 1 0^{- 3}$

$6.8 \times 1 0^{- 4}$

$6.8 \times 1 0^{- 3}$

Q. Ionisation constant of CH3​COOH is 1.7×10−5 and concentration of H+ions is 3.4×10−4. Then, find out initial concentration of CH3​COOH molecules.

3.4×10−4

3.4×10−3

6.8×10−4

6.8×10−3

Q. Ionisation constant of $C H_{3} COO H$ is $1.7 \times 1 0^{- 5}$ and concentration of $H^{+}$ ions is $3.4 \times 1 0^{- 4}$ . Then, find out initial concentration of $C H_{3} COO H$ molecules.

$3.4 \times 1 0^{- 4}$

$3.4 \times 1 0^{- 3}$

$6.8 \times 1 0^{- 4}$

$6.8 \times 1 0^{- 3}$