# Calculate the pH of a buffered solution after adding strong acid [closed]

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. :)

• We will not help you solve this problem if you don't send us what sort of trials and calculations you have done previously. Aug 5 at 10:07
• 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. Aug 5 at 10:12
• there is a general method outlined here chemistry.stackexchange.com/questions/60068/… Aug 5 at 10:17
• Adding 0.2 millimole HCl to the buffer solution is consuming 0.2 millimole CH3NH2 and creating 0.2 millimole CH3NH3Cl. Go on ! Aug 5 at 10:19
• @Maurice Thanks! Can you take a quick look at my solution? Aug 5 at 12:38

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?

• 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?
– Karsten
Aug 5 at 12:42
• @Karsten. Why do you ask this question ? Why transform the initial question ? What does it help ? Aug 5 at 12:46
• @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.
– Karsten
Aug 5 at 12:50
• @Karsten. OK. I see. Thank you for the information. I was also surprised by the self-answer. Aug 5 at 15:34