-2
$\begingroup$

Calculate the pH of a solution formed by the addition of $0.0020$ moles of $\ce{HCl}$ to $\pu{100.0 mL}$ of $\pu{0.200 M}$ $\ce{CH3NH2}$/ $\ce{0.100 M CH3NH3Cl}$.

$K_\mathrm{b}(\ce{CH3NH2})=4.4\times 10^{-4}$

Any hints will be much appreciated. :)

$\endgroup$
6
  • $\begingroup$ We will not help you solve this problem if you don't send us what sort of trials and calculations you have done previously. $\endgroup$
    – Maurice
    Aug 5, 2023 at 10:07
  • $\begingroup$ I know this is a buffer solution that resists a change in pH. I can calculate the pH of the buffer by itself, but how do I proceed after the HCl is added? I just want a nudge in the right direction. $\endgroup$
    – Shoes
    Aug 5, 2023 at 10:12
  • 1
    $\begingroup$ there is a general method outlined here chemistry.stackexchange.com/questions/60068/… $\endgroup$
    – porphyrin
    Aug 5, 2023 at 10:17
  • 1
    $\begingroup$ Adding 0.2 millimole HCl to the buffer solution is consuming 0.2 millimole CH3NH2 and creating 0.2 millimole CH3NH3Cl. Go on ! $\endgroup$
    – Maurice
    Aug 5, 2023 at 10:19
  • $\begingroup$ @Maurice Thanks! Can you take a quick look at my solution? $\endgroup$
    – Shoes
    Aug 5, 2023 at 12:38

1 Answer 1

1
$\begingroup$

Okay, here goes nothing.

$$\ce{CH3NH3+ <=> H+ + CH3NH2}$$

$$K_a(\ce{CH3NH3+})=\frac{[\ce{H+}][\ce{CH3NH2}]}{[\ce{CH3NH3+}]}\tag{1}$$

$\displaystyle K_a(\ce{CH3NH3+})=\frac{10^{-14}}{K_b(\ce{CH3NH2})}=\frac{10^{-14}}{4.4\times 10^{-4}}=2.27\times 10^{-11}$

Initial moles of $\ce{CH3NH3+}=0.100\times 0.100=0.010$

Initial moles of $\ce{CH3NH2}=0.100\times 0.200=0.020$

$$\ce{H+ + CH3NH2 -> CH3NH3+}$$

$0.0020$ moles of $\ce{H+}$ react with $0.0020$ moles of $\ce{CH3NH2}$ to form $0.0020$ moles of $\ce{CH3NH3+}$.

Final moles of $\ce{CH3NH3+}=0.010+0.0020=0.012$

Final moles of $\ce{CH3NH2}=0.020-0.0020=0.018$

$V=100.0\text{ mL}=0.1000\text{ L}$

$\displaystyle[\ce{CH3NH3+}]_f=\frac{0.012}{0.1000}=0.120$

$\displaystyle[\ce{CH3NH2}]_f=\frac{0.018}{0.1000}=0.180$

Substituting these values in equation $(1)$, we get

$\displaystyle[\ce{H+}]=\frac{2.27\times 10^{-11}\times[\ce{CH3NH3+}]}{[\ce{CH3NH2}]}=\frac{2.27\times 10^{-11}\times 0.120}{0.180}=1.51\times 10^{-11}$

pH$=-\log[\ce{H+}]=10.82$

Does this look good?

$\endgroup$
4
  • 1
    $\begingroup$ What was the information you needed when you posted the question? Which comment did it? What would happen if you added 0.03 moles of strong acid instead? $\endgroup$
    – Karsten
    Aug 5, 2023 at 12:42
  • 1
    $\begingroup$ @Karsten. Why do you ask this question ? Why transform the initial question ? What does it help ? $\endgroup$
    – Maurice
    Aug 5, 2023 at 12:46
  • 1
    $\begingroup$ @Maurice I was surprised by the self-answer and was wondering what the missing step was initially. My 3rd question in the comment goes after limits of the recipe shown here. $\endgroup$
    – Karsten
    Aug 5, 2023 at 12:50
  • 1
    $\begingroup$ @Karsten. OK. I see. Thank you for the information. I was also surprised by the self-answer. $\endgroup$
    – Maurice
    Aug 5, 2023 at 15:34

Not the answer you're looking for? Browse other questions tagged or ask your own question.