Question

A buffer solution is made by dissolving 0.45 moles of a weak acid (HA) and 0.33 moles of KOH into 710 mL of solution. What is the pH of this buffer? Ka = 6x10-6 for HA.

Answer 1

Answer: The pH of the buffer solution is 5.08.

Solution:

$K_a\text{of the acid HA}=6\times 10^{-6}$

Volume of buffer solution=710 mL = 0.710 L (1L=1000ml)

$[concentration]=\frac{\text{number of moles}}{\text{volume in litres}}$

$[HA]=\frac{0.45}{0.710} mol/L,[KOH]=\frac{0.33}{0.710} mol/L$

ph of the buffer can be calculated by using buffer equation:

$pH=pK_a+log\frac{[base]}{[acid]}$...(1)

On substituting values in (1)  we get

$pH=-log(6\times 10^{-6})+log\frac{\frac{0.33}{0.710}}{\frac{0.45}{0.710}}=5.08$

The pH of the buffer solution is 5.08.

Answer 2

5.66

Further explanation

Given:

• 0.45 moles of a weak acid (HA) and 0.33 moles of KOH
• A buffer solution is made by dissolving a weak acid (HA) and KOH into 710 mL of solution.
• $\boxed{ \ K_a = 6 \times 10^{-6} \ for \ HA \ }$

Question:

What is the pH of this buffer?

The Process:

The balanced equation for this reaction is:

              $\boxed{ \ HA + KOH \rightarrow KA + H_2O \ }$

Initial:      0.45     0.33         -          -

Change:  0.33     0.33     0.33    0.33

Final:       0.12         -        0.33    0.33

          (excess)  (limiting)

When 0.45 moles of HA react with 0.33 moles of KOH, 0.33 moles of KA are produced, and 0.12 moles of HA remain unreacted.

At the end of the above reaction, the buffer solution (or buffer system) is composed of weak acids mixed with their conjugate base salts.

To calculate the specific pH of a given buffer, we need using The Henderson-Hasselbalch equation for acidic buffers:

$\boxed{ \ pH = pKa + log_{10} \frac{[A-]}{[HA]} \ }$

where,

• Ka represents the dissociation constant for the weak acid;
• [A-] represent the concentration of the conjugate base (i.e. salt);
• [HA] is the concentration of the weak acid.

Let us prepare the concentrations.

$\boxed{ \ [HA] = \frac{0.12 \ moles}{0.710 \ L} = 0.169 \ M \ }$

From KA salt ionization:  $\boxed{ \ KA \rightarrow K^+ + A^-\ }$ we observe that there is one A⁻ ion or the valence is 1.

Hence, $\boxed{ \ [A^-] = [KA] \times valence \ }$, so

$\boxed{ \ [A^-] = \frac{0.33 \ moles}{0.710 \ L} \times 1 = 0.465 \ M \ }$

Let us calculate the pH of this buffer.

$\boxed{ \ pH = -log(6 \times 10^{-6}) + log \Big( \frac{0.465}{0.169} \Big) \ }$

Recall this: $\boxed{ \ - log(a \times 10^b) \rightleftharpoons b - log \ a \ }$

$\boxed{ \ pH = 6 - log 6 + log \Big( \frac{0.465}{0.169} \Big) \ }$

Thus, the pH of this buffer is 5.66 and represents an acidic buffer solution.

_ _ _ _ _ _ _ _ _ _

Notes:

A buffer solution is one that naturally resists changes in pH when lesser quantities of an acid or an alkali are carefully added to it.  

• Acidic buffer solutions. An acidic buffer solution is simply one which has a pH of less than 7. Acidic buffer solutions are commonly gained from a weak acid and one of its salts.  
• Alkaline buffer solutions. An alkaline buffer solution has a pH greater than 7. Alkaline buffer solutions are commonly gained from a weak base and one of its salts.

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