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Anastasy [175]
3 years ago
10

A student performs a titration procedure by adding NaOH to a solution that contains H+ ions. She added the indicator, phenolphth

alein, to the flask containing the H+ ions. What was the purpose of adding the indicator to the flask?
A. The indicator adds more H+ ions to the solution which allows the titration procedure to proceed faster.


B. The indicator is expected to turn pink when the end point of the neutralization reaction occurs.


C. The indicator was used to determine when the amount of H+ ions was greater than the concentration of OH- ions.


D. Because the indicator has the same pH as the NaOH, it is used to determine the pH of the titrant.
Chemistry
1 answer:
Jobisdone [24]3 years ago
7 0
Indicators are substances that produce colour changes to indicate the presence or absence of a threshold concentration of a particular chemical specie. In neutralization reaction, indicators produce colour changes on the basis of the pH of the solution. In acid base titration, indicators are used to indicate the end point of the neutralization reaction. The end point of acid base titration is the equivalence point when the moles of the of the standard solution is equal to the moles of the unknown solution. Thus, the correct option is B. Note that the solution that contain the H+ ions is the acid.
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5 0
3 years ago
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
If 155 grams of potassium (K) reacts with 122 grams of potassium nitrate (KNO3), what is the limiting reagent?
GenaCL600 [577]
K:

m=155g
M=39g/mol

n = 155g / 39g/mol ≈ 3,97mol

KNO₃:

m=122g
M=101g/mol

n = 122g/101g/mol = 1,21mol

2K          +            10KNO₃  ⇒  6K₂O + N₂
2mol        :            10mol
3,97mol   :           1,21mol
                             limiting reagent

KNO₃ is limiting reagent

5 0
3 years ago
State with reasons, whether sulphur dioxide is acting as an oxidizing agent or a reducing agent in each of the following reactio
evablogger [386]

Answer:

A) oxidizing agent is SO2

B) NaClO is the oxidizing agent

Explanation:

A) This is a redox reaction in which oxidation and reduction occur simultaneously.

Thus, in 2H2S(g) + SO2(g) -> 2H2O(l) + 3S(s);

H2S is reduced as follows;

H2S → S + 2H+ + 2e−

We can see that SO2 has been reduced while H2S gets oxidized since it has changed state from - 2 to 0 . Thus sulphur dioxide is the oxidizing agent.

B) SO2(g) + H2O(l) + NaClO(aq) -> NaCl(aq) + H2SO4(aq)

In this, SO2 undergoes oxidation and NaClO is the oxidizing agent

4 0
3 years ago
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