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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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PLZZZZ HELP!!!!!! WILL GIVE BRAINLIEST!!
guapka [62]
Color change bubbles and i think dissolving
4 0
3 years ago
Read 2 more answers
To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?
poizon [28]
We know that to relate solutions of with the factors of molarity and volume, we can use the equation: M_{1}  V_{1} = M_{2}  V_{2}

**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.

So now we can assign values to these variables. Let us say that the 18 M H_{2}  SO_{4} is the left side of the equation. Then we have:

(18 M)(0.050 L)=(4.35M) V_{2}

We can then solve for V_{2}:

V_{2}= \frac{(18M)(0.05L)}{4.35M} and V_{2} =0.21 L or 210 mL

We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.


7 0
3 years ago
(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S
Sunny_sXe [5.5K]

These are two questions and two answers

Answer:

    Question 1:

  • <u>H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)</u>

    Question 2:

  • <u>0.201 M</u>

Explanation:

<u>Question 1:</u>

The<em> neutralization</em> reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) <u>Word equation:</u>

  • sulfuric acid + potassium hydroxide → potassium sulfate + water

                 ↑                               ↑                              ↑                       ↑

               acid                          base                        salt                   water

<u>2) Skeleton equation (unbalanced)</u>

  • H₂SO₄ + KOH → K₂SO₄ + H₂O

<u>#) Balanced chemical equation (including phases)</u>

  • H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

<u>Question 2:</u>

<u>1) Mol ratio:</u>

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

  • 1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

<u>2) Moles of H₂SO₄:</u>

  • V = 0.750 liter
  • M = 0.480 mol/liter
  • M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

<u>3) Moles of KOH:</u>

  • V = 0.700 liter
  • M = 0.290 mol/liter
  • M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

<u>4) Determine the limiting reagent:</u>

a) Stoichiometric ratio:

   1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

   0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄  is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

<u>5) Amount of H₂SO₄ that reacts:</u>

  • Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:

         x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

         x = 0.203 / 2 = 0.0677 mol of H₂SO₄

<u>6) Concentration of H₂SO₄ remaining:</u>

  • Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol

  • Total volume = 0.700 liter + 0.750 liter = 1.450 liter

  • Concetration = M

        M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

6 0
3 years ago
Will name brainliest!!!!!
dem82 [27]

According to sources, the most probable answer to this query is the reaction is spontaneous and will proceed without any energy input. This means that energy was produced and can result to a reaction.We appreciate your questions. Please, never hesitate to ask more in Brainly your queries. 
7 0
3 years ago
Read 2 more answers
which nonmetals could form an ionic compound with the formula MgX^2 ( where X represents the nonmetal)? (possible, not possible)
djverab [1.8K]

Answer:

The answer to your question is:

Explanation:

Ionic compound is formed when a metal and a nonmetal are attached.

If we have MgX₂, that means that the nonmetal must have a valence of -1.

From the list the nonmetals with a valence of -1 are:

Bromine(Br) and  fluorine(F).

7 0
3 years ago
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