mathrmCH3 mathrmNH2(0.12. mole, . mathrmpK mathrmb=3.3) is added to 0.08 moles of HCl and the solution is diluted to one litre, resulting pH of solution is:

Question

mathrmCH3 mathrmNH2(0.12. mole, . mathrmpK mathrmb=3.3) is added to 0.08 moles of HCl and the solution is diluted to one litre, resulting pH of solution is: (A) 10.7 (B) 3.6 (C) 10.4 (D) 11.3

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-4ljgb11sgyew.png" /> pOH=pKb+log([C H3 N H3+]/[(C H3 N H2)]) =3.3+log(0.08/0.04)=3.6 ∴ pH=10.4

Q. $CH_{3} NH_{2} (0.12$ mole, $pK_{b} = 3.3)$ is added to $0.08$ $m o l es$ of $H Cl$ and the solution is diluted to one litre, resulting $p H$ of solution is:

$10.7$

$3.6$

$10.4$

$11.3$

$pO H = p K_{b} + l o g \frac{[ C H _{3} N H _{3}^{+} ]}{[ ( C H _{3} N H _{2} ) ]}$

$= 3.3 + l o g \frac{0.08}{0.04} = 3.6$

$∴ p H = 10.4$

Q. CH3​NH2​(0.12 mole, pKb​=3.3) is added to 0.08 moles of HCl and the solution is diluted to one litre, resulting pH of solution is:

10.7

3.6

10.4

11.3