When 50 mL of 0.1 M NH3 is mixed with 10 mL of 0.1 M HCl then what is the pH of resultant solution? (pKb = 4.75 )

Question

When 50 mL of 0.1 M NH3 is mixed with 10 mL of 0.1 M HCl then what is the pH of resultant solution? (pKb = 4.75 ) (A) 9.25 (B) 10 (C) 9.85 (D) 4.15

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/a69312a303435518ab136f0717b43b87-.png" /> pOH=pKb+log ([salt]/[base]) =4.75+log((1/4))=4.15 pH=14-pOH=14-4.15=9.85

Q. When $50 m L$ of $0.1 MN H_{3}$ is mixed with $10 m L$ of $0.1 M H Cl$ then what is the pH of resultant solution? $(p K_{b} = 4.75$ )

9.25

10

9.85

4.15

$pO H = p K_{b} + l o g \frac{[ s a lt ]}{[ ba se ]}$
$= 4.75 + l o g (\frac{1}{4}) = 4.15$
$p H = 14 - pO H = 14 - 4.15 = 9.85$

Q. When 50mL of 0.1MNH3​ is mixed with 10mL of 0.1MHCl then what is the pH of resultant solution? (pKb​=4.75 )

9.25

10

9.85

4.15

pOH=pKb​+log[base][salt]​ =4.75+log(41​)=4.15 pH=14−pOH=14−4.15=9.85

Q. When $50 m L$ of $0.1 MN H_{3}$ is mixed with $10 m L$ of $0.1 M H Cl$ then what is the pH of resultant solution? $(p K_{b} = 4.75$ )

$pO H = p K_{b} + l o g \frac{[ s a lt ]}{[ ba se ]}$
$= 4.75 + l o g (\frac{1}{4}) = 4.15$
$p H = 14 - pO H = 14 - 4.15 = 9.85$