The Henderson's equation for acetic acid and sodium acetate buffer is given by the expression

Question

The Henderson's equation for acetic acid and sodium acetate buffer is given by the expression (A) PH = p Ka- log ([ CH 3 COONa ]/[ CH 3 COOH ]) (B) pH = p Ka+ log ([ CH 3 COONa ]/[ CH 3 COOH ]) (C) pOH = pK a- log ([ CH 3 COONa ]/[ CH 3 COOH ]) (D) . pOH = p Ka- log ([ CH 3 COOH ]/[ CH 3 COONa .)]

Tardigrade Admin · Accepted Answer

Correct answer is (b) pH = p Ka+ log ([ CH 3 COONa ]/[ CH 3 COOH ])

Q. The Henderson's equation for acetic acid and sodium acetate buffer is given by the expression

$P H = p K_{a} - lo g \frac{[ C H _{3} COON a ]}{[ C H _{3} COO H ]}$

$p H = p K_{a} + lo g \frac{[ C H _{3} COON a ]}{[ C H _{3} COO H ]}$

$pO H = p K_{a} - lo g \frac{[ C H _{3} COON a ]}{[ C H _{3} COO H ]}$

$pO H = p K_{a} - lo g \frac{[ C H _{3} COO H ]}{[ C H _{3} COON a}]$

Correct answer is (b) $p H = p K_{a} + lo g \frac{[ C H _{3} COON a ]}{[ C H _{3} COO H ]}$

Q. The Henderson's equation for acetic acid and sodium acetate buffer is given by the expression

PH=pKa​−log[CH3​COOH][CH3​COONa]​

pH=pKa​+log[CH3​COOH][CH3​COONa]​

pOH=pKa​−log[CH3​COOH][CH3​COONa]​

pOH=pKa​−log[CH3​COONa[CH3​COOH]​]

Correct answer is (b) pH=pKa​+log[CH3​COOH][CH3​COONa]​

$P H = p K_{a} - lo g \frac{[ C H _{3} COON a ]}{[ C H _{3} COO H ]}$

$p H = p K_{a} + lo g \frac{[ C H _{3} COON a ]}{[ C H _{3} COO H ]}$

$pO H = p K_{a} - lo g \frac{[ C H _{3} COON a ]}{[ C H _{3} COO H ]}$

$pO H = p K_{a} - lo g \frac{[ C H _{3} COO H ]}{[ C H _{3} COON a}]$

Correct answer is (b) $p H = p K_{a} + lo g \frac{[ C H _{3} COON a ]}{[ C H _{3} COO H ]}$