pH or 1L 0.1 M HCl when 5 g of HF is added:

Question

pH or 1L 0.1 M HCl when 5 g of HF is added: (A) 0.53 (B) 1 (C) 0.1 (D) 0.01

Tardigrade Admin · Accepted Answer

pH of mixture of strong and weak acid depends only upon strong acid concentration. 0.1 M HCl=0.1 NHCl its [H+]=0.1 p H=- log [H+]= log (1/[H+]) = log (1/[0.1]) = log 10=1

Q. $p H$ or $1 L 0.1 M H Cl$ when $5 g$ of $H F$ is added:

0.53

1

0.1

0.01

$p H$ of mixture of strong and weak acid depends only upon strong acid concentration.
$0.1 M H Cl = 0.1 N H Cl$ its $[H^{+}] = 0.1$
$p H = - lo g [H^{+}] = lo g \frac{1}{[ H ^{+} ]}$
$= lo g \frac{1}{[ 0.1 ]} = lo g 10 = 1$

Q. pH or 1L0.1MHCl when 5g of HF is added:

0.53

1

0.1

0.01

pH of mixture of strong and weak acid depends only upon strong acid concentration. 0.1MHCl=0.1NHCl its [H+]=0.1 pH=−log[H+]=log[H+]1​ =log[0.1]1​=log10=1

Q. $p H$ or $1 L 0.1 M H Cl$ when $5 g$ of $H F$ is added:

$p H$ of mixture of strong and weak acid depends only upon strong acid concentration.
$0.1 M H Cl = 0.1 N H Cl$ its $[H^{+}] = 0.1$
$p H = - lo g [H^{+}] = lo g \frac{1}{[ H ^{+} ]}$
$= lo g \frac{1}{[ 0.1 ]} = lo g 10 = 1$