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Describe the pH of the solutions given the indicator colors and arrange them by acidity. Also, list indicator colors in different pH solutions.

Chemistry

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Answer

d

2.0×10−4 M\text{2.0} \times 10^{-4} \text{ M}

Step-by-step solution

4 steps

Part a· 1 step

Step 1: Identify the acidity/basicity of each solution based on the indicator's color.

The indicator turns orange in acidic, greenish in neutral, and bluish in alkaline solutions.

  • Glucose: Green →\rightarrow Neutral
  • Rain water: Pale or Orange →\rightarrow Slightly acidic to acidic
  • Lime water: Blue →\rightarrow Basic
  • Vinegar: Orange →\rightarrow Acidic
  • Stomach powder: Pale Blue →\rightarrow Slightly basic
  1. Arrange the solutions in increasing order of acidity (from most basic to most acidic).

    1. Lime water (Blue - most basic)
    2. Stomach powder (Pale Blue - slightly basic)
    3. Glucose (Green - neutral)
    4. Rain water (Pale or Orange - slightly acidic)
    5. Vinegar (Orange - most acidic)

    The order of increasing acidity is:
    Lime water, Stomach powder, Glucose, Rain water, Vinegar.

Part b

Here are the colors of the specified indicators in neutral, base, and acid solutions:

IndicatorsNeutral solutionBase solutionAcid solution
Methyl orangeOrangeYellowRed
Methyl greenGreenBlueYellow
PhenolphthaleinColorlessPinkColorless
Thymol blueYellowBlueYellow
Bromothymol blueGreenBlueYellow
Methyl redOrangeYellowRed
Cresol redYellow/OrangeVioletYellow
LitmusPurpleBlueRed

Part c

Here are suitable indicators for the given types of titrations:

Analytes SuitableIndicators
Weak acid / Weak BaseNo suitable indicator (potentiometric titration preferred)
Strong acid / Strong BaseBromothymol blue, Litmus
Strong acid / Weak BaseMethyl orange, Methyl red
Strong base / Weak acidPhenolphthalein, Thymol blue

Part d

2.0×10−4 M\text{2.0} \times 10^{-4} \text{ M}

Step 1: Write the ionization equilibrium for the indicator (HIn) and its ionization constant (KaK_a) expression.

The ionization of a weak acid indicator is:

HIn⇌H++In−\text{HIn} \rightleftharpoons \text{H}^+ + \text{In}^-

The ionization constant (KaK_a) is given by:

Ka=[H+][In−][HIn]K_a = \frac{[\text{H}^+][\text{In}^-]}{[\text{HIn}]}
  1. Rearrange the KaK_a expression to solve for [H+][\text{H}^+].

    [H+]=Ka×[HIn][In−][\text{H}^+] = K_a \times \frac{[\text{HIn}]}{[\text{In}^-]}
  2. Substitute the given values into the equation.

    Given:
    Ka=10−5K_a = 10^{-5}
    [In−]=0.03 M[\text{In}^-] = 0.03 \text{ M}
    [HIn]=0.61 M[\text{HIn}] = 0.61 \text{ M}

    [H+]=(10−5)×0.61 M0.03 M[\text{H}^+] = (10^{-5}) \times \frac{0.61 \text{ M}}{0.03 \text{ M}}
  3. Calculate the concentration of hydrogen ion.

    [H+]=10−5×20.333...[\text{H}^+] = 10^{-5} \times 20.333... [H+]=2.0333...×10−4 M[\text{H}^+] = 2.0333... \times 10^{-4} \text{ M}

    Rounding to two significant figures:

    [H+]=2.0×10−4 M[\text{H}^+] = \boxed{\text{2.0} \times 10^{-4} \text{ M}}
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