In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the con

Q: Why do we use the Henderson-Hasselbalch equation here?

Because it directly relates the pH of a buffer solution to the $pK_a$ of the weak acid and the ratio of the concentrations of the salt and the acid. It is the standard formula for buffer calculations.

Q: What if the buffer was basic?

For a basic buffer (a weak base and its salt), we would use the counterpart equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$, and then find the pH using the relation $pH = 14 - pOH$.

Q: Why is the concentration of the salt so much higher than the acid?

The desired pH ($5.74$) is exactly 1 unit higher than the $pK_a$ ($4.74$). Because the Henderson-Hasselbalch equation uses a base-10 logarithm, a difference of 1 in pH means the ratio of salt to acid must be $10^1$, which is a 10:1 ratio.

Q: Does the volume of the solution matter here?

The Henderson-Hasselbalch equation uses the ratio of concentrations. Since both the acid and salt are dissolved in the same total volume of the buffer solution, the volume mathematically cancels out. You could also use the ratio of their moles directly.

Q: What exactly is the significance of $pK_a$?

$pK_a$ is the negative base-10 logarithm of the acid dissociation constant ($K_a$). It is a measure of the strength of the weak acid. A lower $pK_a$ means a relatively stronger weak acid. When the concentrations of salt and acid are equal, the $\log(1)$ term becomes zero, and the pH of the buffer exactly equals the $pK_a$.

Question

In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is ………… M. (Round off to the nearest integer).
[Given : $pK_a$ (acetic acid) = 4.74]

Sigma Percentile · Accepted Answer

### The Magic of Buffers

Imagine a chemical shock absorber. You add a splash of strong acid or a dash of strong base to a solution, and instead of the pH swinging wildly, it barely budges. This is the magic of a **buffer solution**.

In our problem, we are dealing with an **acidic buffer**. It is crafted by mixing a weak acid—in this case, acetic acid ($CH_3COOH$)—with its conjugate base, provided by a salt like sodium acetate ($CH_3COONa$). The weak acid stands ready to neutralize any incoming base, while the conjugate base is on guard to mop up any incoming acid. Together, they hold the pH steady.

### The Master Key: Henderson-Hasselbalch Equation

To control this chemical dance and set the pH exactly where we want it, we use the **Henderson-Hasselbalch equation**. It is the ultimate tool for buffer calculations, elegantly linking the pH of the solution to the intrinsic strength of the acid ($pK_a$) and the ratio of the salt and acid concentrations.

For an acidic buffer, the equation is:

$$pH = pK_a + \log_{10} \left( \frac{[	ext{Salt}]}{[	ext{Acid}]} ight)$$

Here, $[	ext{Salt}]$ represents the concentration of the conjugate base ($CH_3COO^-$ from sodium acetate), and $[	ext{Acid}]$ is the concentration of the weak acid ($CH_3COOH$).

### Crunching the Numbers

We are given a target pH of $5.74$. The $pK_a$ of acetic acid is a known constant, $4.74$. We also know the concentration of the acetic acid in our beaker is exactly $1.0 	ext{ M}$. Our mission is to find the missing piece of the puzzle: the concentration of sodium acetate.

Let's substitute our known values into the master equation:

$$5.74 = 4.74 + \log_{10} \left( \frac{[CH_3COONa]}{1.0} ight)$$

To isolate the logarithmic term, we simply subtract $4.74$ from both sides. Notice how beautifully the numbers align:

$$5.74 - 4.74 = 1.0$$

So, our equation simplifies to:

$$1.0 = \log_{10} ([CH_3COONa])$$

### The Power of Ten

We are almost there. To free the concentration from the logarithm, we need to convert the equation from its logarithmic form to its exponential form. Remember the fundamental rule of logarithms: if $\log_{10}(x) = y$, then $x = 10^y$.

Applying this rule, we take the antilog of both sides:

$$[CH_3COONa] = 10^{1.0}$$

$$[CH_3COONa] = 10 	ext{ M}$$

And there is our answer! To push the pH exactly one unit above the $pK_a$ (from $4.74$ to $5.74$), we need the salt concentration to be exactly ten times the acid concentration. Since the acid is $1.0 	ext{ M}$, the salt must be a whopping **$10 	ext{ M}$**. The nearest integer is simply $10$.

Enter Numerical Value:

Visualized Solution

\text{Acidic Buffer System}

\text{Henderson-Hasselbalch Equation}

\text{Substituting the Values}

\text{Isolating the Logarithmic Term}

\text{Removing the Logarithm}

\text{Final Result}

\text{Food for Thought}

The Sigma Insight: pH, Buffer and Indicator

The Magic of Buffers

The Master Key

Crunching the Numbers

The Power of Ten

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