( textCH)3 ( textNH)2 (0.12 textmole , ( textpK) textb = 3.3) is added to 0.08 moles of HCl and the solution is diluted to one litre, resulting pH of solution is:

Question

( textCH)3 ( textNH)2 (0.12 textmole , ( textpK) textb = 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-o6fy1wms6r4gt0lf.jpg" /> 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. (CH)3​(NH)2​(0.12mole,(pK)b​=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