Question

A weak acid type indicator was found to be 60%-dissociated at $p^H=9$. find % D.O.D at $p^H=8$

Answer 1

13.04%

Answer 2

The indicator is a weak acid, HIn, which dissociates as follows:

HIn (aq) $\rightleftharpoons$ H$^+$ (aq) + In$^-$ (aq)

The dissociation constant ($K_{In}$) or its negative logarithm ($pK_{In}$) for the indicator is related to the pH and the ratio of the conjugate base form ($In^-$) to the acid form ($HIn$) by the Henderson-Hasselbalch equation:

$pH = pK_{In} + \log \frac{[In^-]}{[HIn]}$

Step 1: Calculate $pK_{In}$ from the given information.

At $pH = 9$, the indicator is 60% dissociated. This means that if we consider the total initial concentration of the indicator to be $C$, then at equilibrium:

Concentration of dissociated form $[In^-]$ = $0.60C$

Concentration of undissociated form $[HIn]$ = $C - 0.60C = 0.40C$

Therefore, the ratio $\frac{[In^-]}{[HIn]} = \frac{0.60C}{0.40C} = \frac{60}{40} = \frac{3}{2} = 1.5$.

Substitute these values into the Henderson-Hasselbalch equation:

$9 = pK_{In} + \log(1.5)$

We know that $\log(1.5) \approx 0.176$.

So, $9 = pK_{In} + 0.176$

$pK_{In} = 9 - 0.176 = 8.824$

Step 2: Calculate the percentage dissociation (D.O.D) at $pH = 8$.

Let $\alpha$ be the degree of dissociation (D.O.D) at $pH = 8$.

Then, the ratio $\frac{[In^-]}{[HIn]} = \frac{\alpha C}{(1-\alpha)C} = \frac{\alpha}{1-\alpha}$.

Substitute the new pH and the calculated $pK_{In}$ into the Henderson-Hasselbalch equation:

$8 = 8.824 + \log \left(\frac{\alpha}{1-\alpha}\right)$

Rearrange to solve for the log term:

$\log \left(\frac{\alpha}{1-\alpha}\right) = 8 - 8.824 = -0.824$

To find the ratio $\frac{\alpha}{1-\alpha}$, take the antilog of both sides:

$\frac{\alpha}{1-\alpha} = 10^{-0.824}$

To calculate $10^{-0.824}$:

$10^{-0.824} = 10^{(-1 + 0.176)}$ $= 10^{-1} \times 10^{0.176}$

Since $\log(1.5) \approx 0.176$, then $10^{0.176} \approx 1.5$.

So, $\frac{\alpha}{1-\alpha} \approx 0.1 \times 1.5 = 0.15$

Now, solve for $\alpha$:

$\alpha = 0.15 (1-\alpha)$

$\alpha = 0.15 - 0.15\alpha$

$\alpha + 0.15\alpha = 0.15$

$1.15\alpha = 0.15$

$\alpha = \frac{0.15}{1.15} = \frac{15}{115} = \frac{3}{23}$

Finally, convert the degree of dissociation to percentage dissociation:

Percentage D.O.D = $\alpha \times 100\%$

Percentage D.O.D = $\frac{3}{23} \times 100\%$

Percentage D.O.D = $\frac{300}{23}\%$

Percentage D.O.D $\approx 13.04\%$