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Vesnalui [34]
3 years ago
13

What are the limitations of litmus paper and Phenolphthalein indicators? Name to other indicators that can be used that do not h

ave such limitations

Chemistry
1 answer:
jenyasd209 [6]3 years ago
5 0

Answer:

Here's what I find.

Explanation:

An indicator is usually is a weak acid in which the acid and base forms have different colours. Most indicators change colour over a narrow pH range.

(a) Litmus

Litmus is red in acid (< pH 5) and blue in base (> pH 8).

This is a rather wide pH range, so litmus is not much good in titrations.

However, the range is which it changes colour includes pH 7 (neutral), so it is good for distinguishing between acids and bases.

(b) Phenolphthalein

Phenolphthalein  is colourless in acid (< pH 8.3) and red in base (> pH 10).

This is a narrow pH range, so phenolphthalein is good for titrating acids with strong bases..

However, it can't distinguish between acids and weakly basic solutions.

It would be colourless in a strongly acid solution with pH =1 and in a basic solution with pH = 8.

(c) Other indicators  

Other acid-base indicators have the general limitations as phenolphthalein. Most of them have a small pH range, so they are useful in acid-base titrations.

The only one that could serve as a general acid-base indicator is bromothymol blue, which has a pH range of 6.0 to 7.6.

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A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the
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Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

\displaystyle c = \frac{n}{V},

where

  • c is the concentration of the solute,
  • n is the number of moles of the solute, and
  • V is the volume of the solution.

V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}.

Acetic (ethanoic) acid:

\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}.

\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}.

Hydrochloric acid HCl:

\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}.

\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}.

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be 5.00\times 10^{-4}\;\text{M}.

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be -x\;\text{M}. x > 0.

\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}.

\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}.

Rewrite as a quadratic equation and solve for x:

x\cdot(x + 5.00\times 10^{-4}) = (1.8\times 10^{-5} )\cdot (0.100 - x)

x\approx 0.00111.

The pH of a solution depends on its H⁺ concentration.

At equilibrium

[\text{H}^{+}\;(aq)] = 5.00\times 10^{-4}\;\text{M} + x\;\text{M} = 0.00161\;\text{M}.

\text{pH} = -\log{[\text{H}^{+}]} = 2.79.

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Answer:

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