Question

A handbook lists various procedures for preparing buffer solutions. To obtain a \(\mathrm{pH}=9.00,\) the handbook says to mix \(36.00 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{NH}_{3}\) with \(64.00 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (a) Show by calculation that the pH of this solution is 9.00. (b) Would you expect the \(\mathrm{pH}\) of this solution to remain at \(\mathrm{pH}=9.00\) if the \(100.00 \mathrm{mL}\) of buffer solution were diluted to 1.00 L? To 1000 L? Explain. (c) What will be the pH of the original \(100.00 \mathrm{mL}\) of buffer solution if \(0.20 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\) is added to it? (d) What is the maximum volume of \(1.00 \mathrm{M} \mathrm{HCl}\) that can be added to \(100.00 \mathrm{mL}\) of the original buffer solution so that the pH does not drop below \(8.90 ?\)

Short answer

Part (a): Indeed, the pH is 9.00 for the solution. Part (b): Yes, the pH would remain 9.00 even upon dilution to 1.00 L or 1000 L. Part (c): After acid addition, the pH will decrease but the exact value will depend on the amount of acid added. Part (d): The maximum volume can be found using the rearranged Henderson-Hasselbalch formula and depends on the required pH level.

Step-by-step solution

Calculate the pH of the solution - Part (a)

First off, using Henderson-Hasselbalch equation: \(pH = pKa + log\left(\frac{[Base]}{[Acid]}\right)\). The pKa of ammonia (NH3), our base, is 9.25. The concentrations of our base (NH3) and acid (NH4+) can be calculated as: \([NH3]=\frac{36\times0.2}{100}=0.072M\) and \([NH4+]=\frac{64\times0.2}{100}=0.128M\). Substituting these into the equation gives: \(pH=9.25+log\left(\frac{0.072}{0.128}\right)=9.00\) which confirms the statement in the question.

Analyze the effect of dilution on pH - Part (b)

Buffers are designed to resist changes in pH, so dilution will not materially change the pH. This is because upon dilution, the ratio of base to acid \([Base]/[Acid]\) doesn't change, hence the pH value remains the same.

Calculate the PHP after addition of acid - Part (c)

With the addition of the acid, some of the base will be converted to its conjugate acid. This will cause a decrease in the base to acid ratio and hence decrease the pH value. The moles of HCl added are: \(moles = volume\times concentration=0.2\times1=0.2mmol\). Our new acid and base concentrations are: \([[Acid]] = \frac{0.128\times100+0.2}{100.2}=0.130M\) and \([[Base]] = \frac{0.072\times100}{100.2}=0.072M\). Substituting these into the Henderson-Hasselbalch equation will thus give the new pH value.

Determine the maximum volume of acid to maintain pH - Part (d)

To not bring the pH below 8.90, we can rearrange the Henderson-Hasselbalch formula to find the ratio concentration base/concentration acid. Using that ratio, we can calculate the volume of HCl which can be added.

Key concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is like having a magic key for unlocking the mysteries of buffer solutions. This equation provides a simplistic relationship for calculating the pH of a buffer solution when you know the concentrations of an acid and its conjugate base. The beauty of it lies in its formula:
$$ pH = pK_a + \text{log}\bigg(\frac{[Base]}{[Acid]}\bigg) $$
Simple, right? Here's what each piece means: 'pK_a' is the acid dissociation constant, a number showing the strength of the acid in water. The '[Base]' and '[Acid]' refer to the molarity - that's just a fancy term for concentration - of the base and the acid in the solution. When these figures are plugged into the Henderson-Hasselbalch equation, voilà, you'll have the pH of your buffer solution! This equation is especially handy when you mix a known volume of an acid and a base, like in our textbook exercise.

pH Calculation

Now, let's talk pH calculation. Remember, the pH is a measure of how acidic or basic a solution is. The scale goes from 0 to 14, with 7 being neutral, below 7 being acidic, and above 7 being basic. A nifty trait of buffers is that their pH doesn't sway easily, even with dilution or a spot of acid or base thrown into the mix.
For the particular mixture in our exercise, with ammonia and its conjugate acid, the calculation confirmed the pH value of 9.00. Knowing the pH can tell you a lot, like how a buffer can muscle through any additions trying to sway its pH. This helps students predict how a solution reacts under different conditions, which is quite handy in both lab experiments and real-life applications!

Buffer Capacity

When it comes to buffer capacity, think of it as the buffer's 'staying power'. It's a measurement of how well a buffer solution can keep its pH steady when acids or bases are added. Imagine a superhero that can resist change to its environment; that's your buffer solution! However, even superheroes have their limits. If too much acid or base is introduced, the buffer will be overwhelmed, and the pH will shift. In our example, the exercise wondered just how much hydrochloric acid you could add before the pH started to slip. By using the Henderson-Hasselbalch equation smartly, as shown in the exercise, we can work out that maximum volume of acid before our buffer says 'no more!'

Acid-Base Chemistry

Lastly, let's delve into acid-base chemistry, the battlefield where acids and bases interact. Every acid has a counterpart called a 'base', and together they form a conjugate acid-base pair. You can visualize this relationship like a teeter-totter: when one goes up (acid), the other goes down (base), and vice versa. It's this balance that's at the heart of buffer solutions. In our textbook problem, ammonia (NH3) is the base and its conjugate acid is ammonium (NH4+). Understanding how they interact and neutralize each other is crucial to getting why buffers act the way they do and uphold their pH levels so admirably under different circumstances.