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So I was doing my chemistry assignment and became stuck. Can I get some help?

a) Calculate the pH of a buffer system that contains 0.40 M of NH3(aq) and 0.50 M of NH4Cl(aq) . Note that the Kb value of NH3(aq) is 1.8×10−5.

My ans for delta pH: 9.158362492

b) Determine the change in pH if 2.50mL of 0.100 M HCl is added to 0.040 L of the buffer system described in part a).

**My ans for delta pH: **

c) Determine the change in pH if 2.50mL of 0.100 M NaOH is added to 0.040 L of the buffer system described in part a).

My ans for delta pH:

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  • $\begingroup$ In (b) notice that the problem asks for the change in pH, not the pH itself. $$\Delta \pu{pH} = 4.65 - 4.17 = - 0.48$$ better yet to use two extra significant figures in intermediate calculations and just round the final answer. $$\Delta \pu{pH} = 4.6478 - 4.1734 = - 0.4744 \ce{->[round]} -0.47$$ $\endgroup$
    – MaxW
    Apr 2 '20 at 20:58
  • $\begingroup$ You can find the way to solve this question in the accepted answer here. $\endgroup$ Apr 2 '20 at 21:37
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Your first result is OK.

But :

First, it does not make any sense to give a result with $\ce{10}$ significant figures when the initial data has only $\ce{2}$ significant figures. When a data is given like here $\ce{1.8 x 10^{-5}}$, it means that the author cannot be more precise, and may admit that the exact value of $\ce{K_b}$ is somewhere between $\ce{1.75 10^{-5}}$ and $\ce{1.85 10^{-5}}$. So if you repeat the same calculation with say $\ce{1.75 10^{-5}}$, you find a still "acceptable" value of $\ce{pH = 4.6569}$. This means that you final result can be between $\ce{4.66}$ and $\ce{4.62}$. It can be written $\ce{pH = 4.64 ± 0.02}$. The six next figures do not have any meaning.

For the second problem, you add $\ce{0.00025}$ mole $\ce{HCl}$ to a mixture containing $\ce{0.016}$ mol $\ce{NH_3}$ and $\ce{0.02}$ mol $\ce{NH_4^+}$. So the final concentration of $\ce{NH_3}$ and $\ce{NH_4^+}$ are respectively $\ce{0.391 M}$ and $\ce{0.503 M}$. The log of ratio of these concentration is $\ce{0.251}$. So that the final pH is $\ce{4.49 ± 0.02}$.

Try to do the third calculation by the same approach.

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  • $\begingroup$ LOL - In the olden days you couldn't make such a horrendous mistake using 4 place logarithm tables or a slide rule! $\endgroup$
    – MaxW
    Apr 2 '20 at 20:29
  • $\begingroup$ Thank you. But I am still confused about the last one. Wouldn't it calculate to the same answer as b. $\endgroup$
    – Samaj
    Apr 2 '20 at 20:29
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    $\begingroup$ @Samajm - If you add an strong acid to a buffer the pH has to go down. If you add a strong base to a buffer the pH has to go up. $\endgroup$
    – MaxW
    Apr 2 '20 at 20:31
  • $\begingroup$ I did all the working but I assume my first answer is wrong. $\endgroup$
    – Samaj
    Apr 2 '20 at 21:13
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Ok, let's beat this to death with ICE tables.

(a) Calculate the pH of a buffer system that contains 0.40 M of $\ce{NH3(aq)}$ and 0.50 M of $\ce{NH4Cl(aq)}$ . Note that the $K_\beta$ value of $\ce{NH3(aq)}$ is $1.8\times10^{−5}$.

Some observations:

  • We want the pH, not the pOH.

  • For $\ce{NH4^+}$, $K_\alpha = \dfrac{K_\mathrm{w}}{K_\beta} = \dfrac{1.00\times10^{-14}}{1.8\times10^{−5}} = 5.556\times10^{-10},\quad \mathrm{p}K_\alpha = 9.2553 $

  • Since $\ce{NH4+(aq) > NH3(aq)}$ the solution will be slightly more acidic than the $\mathrm{p}K_\alpha$

  • We'll assume that the equilibrium between $\ce{NH4+(aq)}$ and $\ce{NH3(aq)}$ doesn't shift so that the Henderson-Hasselbalch approximation can be used.

$\begin{array}{|c|c|c|} \hline \ & \ce{NH3} & \ce{NH4^+} \\ \hline \text{I} & \pu{0.400 M} & \pu{0.500 M} \\ \text{C} & 0 & 0 \\ \text{E} & \pu{0.400 M} & \pu{0.500 M} \\ \hline \end{array}\\$

The Henderson-Hasselbalch approximation gives us a method to approximate the pH of the weakly acidic buffer solution as follows:

$$\mathrm{pH} \approx \mathrm{p}K_\mathrm{a} + \log\dfrac{\ce{[NH3]}}{\ce{[NH4+]}}$$ $$ = 9.2553 + \log\dfrac{0.40}{0.50} = 9.1584 \ce{->[Round] = 9.16}$$

(b) Determine the change in pH if 2.50 mL of 0.100 M HCl is added to 0.040 L of the buffer system described in part (a).

Some observations:

  • Again we need to determine the pH, which we can subtract from the pH in part (a) to get the change in pH.

  • Let's just work in millimoles, mM. Since molarity is moles/volume, and the volume for $\ce{NH4+(aq)}$ and $\ce{NH3(aq)}$ is the same, the volume term just cancels. This saves some work calculating the dilutions.

  • HCl is a strong acid and shifts the equilibrium according to the reaction:

$$\ce{NH3 + H+ -> NH4+}$$

  • We'll assume that the equilibrium between $\ce{NH4+(aq)}$ and $\ce{NH3(aq)}$ doesn't shift further so that the Henderson-Hasselbalch approximation can be used.

$\begin{array}{|c|c|c|} \hline \ & \ce{NH3} & \ce{NH4^+} \\ \hline \text{I} & \pu{16 mM} & \pu{20 mM} \\ \text{C} & \pu{-0.25 mM} & \pu{+0.25 mM} \\ \text{E} & \pu{15.75 mM} & \pu{20.25 mM} \\ \hline \end{array}\\$

$$\mathrm{pH} \approx \mathrm{p}K_\mathrm{a} + \log\dfrac{\pu{mM \ce{NH3}}}{\pu{mM \ce{NH4^+}}}$$ $$ = 9.2553 + \log\dfrac{15.75}{20.25} = 9.1462$$

Therefore $\Delta\mathrm{pH} = 9.1462 - 9.1584 = ‭-0.0122‬ \ce{->[round]} -0.01$

(c) Determine the change in pH if 2.50mL of 0.100 M NaOH is added to 0.040 L of the buffer system described in part (a).

Some observations:

  • Again we need to determine the pH, which we can subtract from the pH in part (a) to get the change in pH.

  • Again, let's just work in millimoles, mM.

  • NaOH is a strong base and shifts the equilibrium according to the reaction:

$$\ce{NH4+ + OH- -> NH3 + H2O}$$

  • We'll assume that the equilibrium between $\ce{NH4+(aq)}$ and $\ce{NH3(aq)}$ doesn't shift further so that the Henderson-Hasselbalch approximation can be used.

$\begin{array}{|c|c|c|} \hline \ & \ce{NH3} & \ce{NH4^+} \\ \hline \text{I} & \pu{16 mM} & \pu{20 mM} \\ \text{C} & \pu{+0.25 mM} & \pu{-0.25 mM} \\ \text{E} & \pu{16.25 mM} & \pu{19.75 mM} \\ \hline \end{array}\\$

$$\mathrm{pH} \approx \mathrm{p}K_\mathrm{a} + \log\dfrac{\pu{mM \ce{NH3}}}{\pu{mM \ce{NH4^+}}}$$ $$ = 9.2553 + \log\dfrac{16.25}{19.75} = 9.1706$$

Therefore $\Delta\mathrm{pH} = 9.1706 - 9.1584 = +0.0122‬ \ce{->[round]} +0.01$

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