Ammonium acetate which is 0.01 M , is hydrolysed to 0.001 M concentration. Calculate the change in pH in 0.001 M solution, if initially pH = pKa

Question

Ammonium acetate which is 0.01 M , is hydrolysed to 0.001 M concentration. Calculate the change in pH in 0.001 M solution, if initially pH = pKa  (A)  5  (B)  10  (C)  100  (D)  1

Tardigrade Admin · Accepted Answer

underset0.01 MCH3COONH4 + H2O arrow underset0.001 MCH3COOH + NH4OH pH = p Ka+ log ([ CH 3 COONH 4]/[ CH 3 COOH ]) = p Ka+ log [(0.01/0.001)] = p Ka+ log 10 = p Ka+1 ∴ Change in pH =1

Q. Ammonium acetate which is $0.01 M$ , is hydrolysed to $0.001 M$ concentration. Calculate the change in $p H$ in $0.001 M$ solution, if initially $p H = p K_{a}$

$5$

$10$

$100$

$1$

$0.01 M C H_{3} COON H_{4} + H_{2} O \to 0.001 M C H_{3} COO H + N H_{4} O H$

$p H = p K_{a} + lo g \frac{[ C H _{3} COON H _{4} ]}{[ C H _{3} COO H ]}$

$= p K_{a} + lo g [\frac{0.01}{0.001}]$

$= p K_{a} + lo g 10$

$= p K_{a} + 1$

$∴$ Change in $p H = 1$

Q. Ammonium acetate which is 0.01M , is hydrolysed to 0.001M concentration. Calculate the change in pH in 0.001M solution, if initially pH=pKa​

5

10

100

1

0.01MCH3​COONH4​​+H2​O→0.001MCH3​COOH​+NH4​OH pH=pKa​+log[CH3​COOH][CH3​COONH4​]​ =pKa​+log[0.0010.01​] =pKa​+log10 =pKa​+1 ∴ Change in pH=1

Q. Ammonium acetate which is $0.01 M$ , is hydrolysed to $0.001 M$ concentration. Calculate the change in $p H$ in $0.001 M$ solution, if initially $p H = p K_{a}$

$5$

$10$

$100$

$1$

$0.01 M C H_{3} COON H_{4} + H_{2} O \to 0.001 M C H_{3} COO H + N H_{4} O H$

$p H = p K_{a} + lo g \frac{[ C H _{3} COON H _{4} ]}{[ C H _{3} COO H ]}$

$= p K_{a} + lo g [\frac{0.01}{0.001}]$

$= p K_{a} + lo g 10$

$= p K_{a} + 1$

$∴$ Change in $p H = 1$