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zvonat [6]
2 years ago
14

The [H3O+] in a solution is increased to twice the original concentration. Which change could occur in the pH? 2.0 to 4.0 1.7 to

1.4 5.0 to 2.5 8.5 to 6.5 6.8 to 9.8
Chemistry
2 answers:
Natali [406]2 years ago
8 0
Answer: second option: 1.70 to 1.40

Explanation:

1) The definition and formula of pH is - log [H₃O⁺]

2) So, if the original concentration is x and the increased concentration is 2x, you get:

pHi = - logx

pHf = - log 2x = - log 2 - logx

⇒ pHf - pHi = - log2 - logx - (- logx) =  - log2 ≈ - 0.30

⇒ pHi - pHf =  0.30 This is the pH of the final solution (with double concentration of hydronium ions) is 0.30 points lower than the pH of the initial pH.

3) The only choice that shows a decrease of 0.30 in the pH is the second option: 1.70 to 1.40. So that is the answer.


Lady bird [3.3K]2 years ago
3 0
The change that could occur in the pH in the situation, <span>the [H3O+] in a solution is increased to twice the original concentration, is</span> 1.70 to 1.40
 pH formula  is - log [H₃O⁺]



pHi = - logx
pHf = - log 2x = - log 2 - logx
- log2 ≈ - 0.30
pHi - pHf =  0.30 
So there is a decrease of 0.30.
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A 0.50 M solution of formic acid, HCOOH, has a pH of 2.02. Calculate the percent ionization of HCOOH
kirza4 [7]

Answer is: <span>the percent ionizationof formic acid is 1,82%.
Chemical reaction: HCOOH(aq) </span>⇄ H⁺(aq) + HCOO⁻(aq).<span>
pKa(</span>HCOOH) = 3,77.

Ka(HCOOH) = 1,7·10⁻⁴.

c(HCOOH) = 0,5 M.

<span> [H</span>⁺] = [HCOO⁻] = x; equilibrium concentration.<span>
[HA] = 0,1 M - x.
Ka = [H</span>⁺] · [HCOO⁻] / [HCOOH].<span>
0,00017 = x² / 0,5 M - x.
Solve quadratic equation: x = 0,0091 M.
α = 0,0091 M ÷ 0,5 M · 100% = 1,82%.</span>

8 0
2 years ago
Read 2 more answers
Identify the oxidizing and reducing agents in the following: 2H+(aq) + H2O2(aq) + 2Fe2+(aq) → 2Fe3+(aq) + 2H2O(l)
schepotkina [342]

Answer :  The oxidizing and reducing agents are, H_2O_2 and Fe^{2+}.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The given redox reaction is:

2H^+(aq)+H_2O_2(aq)+2Fe^{2+}(aq)\rightarrow 2Fe^{3+}(aq)+2H_2O(l)

The half oxidation-reduction reactions are:

Oxidation reaction : Fe^{2+}\rightarrow Fe^{3+}+1e^-

Reduction reaction : O^-+1e^-\rightarrow O^{2-}

The oxidation state of oxygen in H_2O_2 and H_2O is, (-1) and (-2) respectively.

In this reaction, 'Fe' is oxidized from oxidation (+2) to (+3) and 'O' is reduced from oxidation state (-1) to (-2). Hence, 'Fe^{2+}' act as a reducing agent and 'H_2O_2' act as a oxidizing agent.

Thus, the oxidizing and reducing agents are, H_2O_2 and Fe^{2+}.

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