Question

Two drops of indicator HIn \(\left(K_{\mathrm{a}}=1.0 \times 10^{-9}\right),\) where HIn is yellow and \(\operatorname{In}^{-}\) is blue, are placed in \(100.0 \mathrm{mL}\) of \(0.10 \mathrm{M}\) HCl. a. What color is the solution initially? b. The solution is titrated with 0.10 \(M\) NaOH. At what pH will the color change (yellow to greenish yellow) occur? c. What color will the solution be after \(200.0 \mathrm{mL}\) NaOH has been added?

Short answer

The initial color of the solution is yellow. The color change (yellow to greenish-yellow) occurs at pH 9. After adding 200.0 mL of 0.10 M NaOH, the color of the solution will be blue.

Step-by-step solution

Determine the initial color of the solution

Since we have \(100.0 \mathrm{mL}\) of \(0.10 \mathrm{M}\) HCl, the initial pH can be calculated as: pH = -log[H3O+], and we know that for HCl, [H3O+] = [HCl], so: pH = -log(0.1) = 1 Since the pH is low (acidic), the indicator HIn (yellow) would dominate. Hence, the initial color of the solution is yellow.

Apply Henderson-Hasselbalch equation to find the pH at which color change occurs

We know that the color change occurs when the concentration of HIn and In- are equal. So, we can use the Henderson-Hasselbalch equation to calculate the pH at that point: pH = pKa + log (Concentration of A-/ Concentration of HA) Since the concentration of HIn and \(\operatorname{In}^-\) are equal, the log term will be zero. Hence, pH = pKa Now, we are given \(K_{a}\) value: \(K_{a} = 1.0 \times 10^{-9}\) To find the pKa, we can use the formula pKa = -log(\(K_{a}\)): pKa = -log(1.0 × 10⁻⁹) = 9 So, the pH at which the color change occurs is 9.

Calculate the color of the solution after adding 200.0 mL of NaOH.

We have added \(200.0 \mathrm{mL}\) of \(0.10 \mathrm{M}\) NaOH, which means that: moles of NaOH added = 0.10 M * 0.200 L = 0.020 mol Since HCl and NaOH react in a 1:1 ratio, we need to find the excess moles of NaOH to determine the final pH of the solution: initial moles of HCl = 0.10 M * 0.100 L = 0.010 mol moles of excess NaOH = moles of NaOH - moles of HCl = 0.020 - 0.010 = 0.010 mol To calculate the final pH, we need to divide the moles of excess NaOH by the total volume of the solution: _total volume = initial volume of HCl solution + volume of NaOH added = 100.0 mL + 200.0 mL = 300.0 mL = 0.3 L _final concentration of OH- = moles of excess NaOH / total volume = 0.010 mol / 0.3 L = 0.0333 M Now we can calculate the pOH and then the pH: pOH = -log[OH-] = -log(0.0333) ≈ 1.48 Since pH + pOH = 14, we can find the pH: pH = 14.00 - 1.48 ≈ 12.52 Since the pH is greater than the pH at which color change occurs (9), the color of the solution will be that of \(\operatorname{In}^-\), which is blue.

Key concepts

Indicator Color Change

Understanding the color change in acid-base titrations is essential for figuring out various stages of the reaction. Indicators are substances that change color based on the pH of the solution they are in. Typically, an indicator has a specific pH range over which it changes color. In our exercise, the indicator HIn is yellow in its acid form and blue in its base form (In-).

Color change occurs when the amount of acid form (HIn) and the base form (In-) are present in approximately equal concentrations. This happens around the pKa value of the indicator, which is an expression of its acid dissociation constant (\( K_a \)). The resulting color can be an intermediate shade, such as greenish-yellow, depending on the indicator and solution pH. It's vital to choose an indicator that changes color close to the equivalence point of the titration to accurately determine when the reaction is complete.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a mathematical relationship used to estimate the pH of a buffer solution. The equation is given by:

\( pH = pKa + \text{log} \frac{[\text{A}^-]}{[\text{HA}]}\text{ , }\)where \( pH \) is the power of hydrogen or the negative logarithm of the hydrogen ion concentration, \( pKa \) is the negative logarithm of the acid dissociation constant (\( K_a \)), \( [\text{A}^-] \) is the concentration of the conjugate base, and \( [\text{HA}] \) is the concentration of the weak acid.

For a weak acid in a solution that also contains its conjugate base, this equation allows us to calculate the pH when we know the concentrations of the acid and base forms. In a titration, the point where the pH equals the pKa of the indicator coincides with the indicator's color change. This is a crucial concept in analyzing titration curves, as it directly relates to measuring the pH of a solution during the titration process.

pH Calculation

The calculation of pH is fundamental in chemistry because it gives us a measure of the acidity or basicity of a solution. The pH scale ranges from 0 to 14, where values less than 7 represent acidic solutions, 7 represents neutrality, and values greater than 7 represent basic solutions.

The pH is calculated using the formula:\[ pH = -\text{log}[\text{H}_3\text{O}^+] \text{ , }\]where \([H3O^+]\) is the concentration of hydronium ions in the solution. For strong acids like HCl, the acid dissociates completely, so the concentration of H+ ions is equal to the concentration of the acid solution. In our titration problem, the initial pH is calculated as such before the addition of any base.

During a titration involving strong acids and bases, once the neutralization is complete, any further addition of base will increase the basicity of the solution, and the pH calculation will require finding the pOH first then converting it to pH using the relation pH + pOH = 14. This knowledge is used to determine the final pH of the solution in the exercise after the titration goes past the equivalence point.

Strong Acid-Strong Base Titration

In a strong acid-strong base titration, a strong base (like NaOH) is gradually added to a solution containing a strong acid (like HCl). They react to form water and a salt in a neutralization reaction that can be represented as: \( HCl + NaOH \rightarrow NaCl + H_2O \). The point at which moles of acid equal moles of base added is known as the equivalence point where the solution is neutral.The titration process is followed using a pH meter or indicator which signifies the end point with a color change. The pH at the equivalence point should theoretically be 7 since the reaction produces neutral water. However, due to subsequent hydrolysis of salts or added excess reagent, the pH may shift.

Our exercise illustrates the scenario after the equivalence point is reached when excess NaOH is added to the HCl solution. This is evident because after adding 200.0 mL of NaOH, the pH of the solution rises above the initial 7 at the equivalence point to a basic pH, which in this case was calculated to be around 12.52, hence turning our indicator blue, indicative of a basic solution.