What will be the pH at the equivalence point during the titration of a 100 mL 0.2 M solution of CH 3 COONa with 0.2 M solution of HCl ? Ka=2 × 10-5

Question

What will be the pH at the equivalence point during the titration of a 100 mL 0.2 M solution of CH 3 COONa with 0.2 M solution of HCl ? Ka=2 × 10-5 (A) 3- log √2 (B) 3+ log √2 (C) 3- log 2 (D) 3+ log 2

Tardigrade Admin · Accepted Answer

CH 3 COONa + HCl longrightarrow CH 3 COOH + NaCl at equivalence the [ CH 3 COOH ]=(20/200) ⇒ 0.1 pH =(1/2) p Ka-(1/2) log C pH =(1/2)[5 log 2- log 10-1] pH =(1/2)[6 log 2] ⇒ pH =3- log √2

Q. What will be the $p H$ at the equivalence point during the titration of a $100 m L 0.2 M$ solution of $C H_{3} COON a$ with $0.2 M$ solution of $H Cl$ ? $K_{a} = 2 \times 1 0^{- 5}$

$3 - lo g 2$

$3 + lo g 2$

$3 - lo g 2$

$3 + lo g 2$

$C H_{3} COON a + H Cl ⟶ C H_{3} COO H + N a Cl$

at equivalence the $[C H_{3} COO H] = \frac{20}{200} \Rightarrow 0.1$

$p H = \frac{1}{2} p K_{a} - \frac{1}{2} lo g C$

$p H = \frac{1}{2} [5 lo g 2 - lo g 1 0^{- 1}]$

$p H = \frac{1}{2} [6 lo g 2] \Rightarrow p H = 3 - lo g 2$

Q. What will be the pH at the equivalence point during the titration of a 100mL0.2M solution of CH3​COONa with 0.2M solution of HCl ? Ka​=2×10−5

3−log2​

3+log2​

3−log2

3+log2

CH3​COONa+HCl⟶CH3​COOH+NaCl at equivalence the [CH3​COOH]=20020​⇒0.1 pH=21​pKa​−21​logC pH=21​[5log2−log10−1] pH=21​[6log2]⇒pH=3−log2​

Q. What will be the $p H$ at the equivalence point during the titration of a $100 m L 0.2 M$ solution of $C H_{3} COON a$ with $0.2 M$ solution of $H Cl$ ? $K_{a} = 2 \times 1 0^{- 5}$

$3 - lo g 2$

$3 + lo g 2$

$3 - lo g 2$

$3 + lo g 2$

$C H_{3} COON a + H Cl ⟶ C H_{3} COO H + N a Cl$

at equivalence the $[C H_{3} COO H] = \frac{20}{200} \Rightarrow 0.1$

$p H = \frac{1}{2} p K_{a} - \frac{1}{2} lo g C$

$p H = \frac{1}{2} [5 lo g 2 - lo g 1 0^{- 1}]$

$p H = \frac{1}{2} [6 lo g 2] \Rightarrow p H = 3 - lo g 2$