Questions: PrairieLearn CHEM 104 ANDINO, F24 Assessments HW15.1. pH Indicators When a solution is titrated with an acid, an indicator (In-) changes from blue to yellow (H In). The Ka of the indicator is 2.40 e-06. At what pH will the indicator color change, first be visible? pH= number (rtol =0.01, atol =1 e-08)

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CHEM 104 ANDINO, F24
Assessments

HW15.1. pH Indicators
When a solution is titrated with an acid, an indicator (In-) changes from blue to yellow (H In). The Ka of the indicator is 2.40 e-06. At what pH will the indicator color change, first be visible?
pH= number (rtol =0.01, atol =1 e-08)

Transcript text: PrairieLearn CHEM 104 ANDINO, F24 Assessments HW15.1. pH Indicators When a solution is titrated with an acid, an indicator ( $\mathrm{In}^{-}$) changes from blue to yellow ( $H I n)$. The $K_{a}$ of the indicator is $2.40 e-06$. At what $p H$ will the indicator color change, first be visible? $p H=$ number (rtol $=0.01$, atol $=1 \mathrm{e}-08$ ) $\square$ Save \& Grade Single attempt Save only Homework 15 Assessment overview Total points: $0 / 1$ Score: $0 \%$ Question HW15.1 Best submission: $0 \%$ All variants: $0 \%$ open Total points: 011 Auto-graded question Report an error in this question Previous question Next question

Solution

Solution Steps

Step 1: Understanding the Problem

We need to determine the pH at which the color change of an indicator becomes visible during a titration. The indicator changes from blue ($\mathrm{In}^{-}$) to yellow ($HIn$), and the given $K_a$ of the indicator is $2.40 \times 10^{-6}$.

Step 2: Using the Henderson-Hasselbalch Equation

The color change of an indicator is typically visible when the concentrations of the acidic form ($HIn$) and the basic form ($\mathrm{In}^{-}$) are equal. This occurs when the pH of the solution is equal to the pKa of the indicator. The pKa is calculated using the formula:

\[ \text{pKa} = -\log(K_a) \]

Step 3: Calculating the pKa

Substitute the given $K_a$ value into the formula:

\[ \text{pKa} = -\log(2.40 \times 10^{-6}) \]

Calculate the pKa:

\[ \text{pKa} \approx 5.6198 \]