Chemistry College

What is the median reaction of second end point in HCL and NaOH titration

Answers

Answer 1

The median reaction at the second end point in the HCl and NaOH titration is: HCl + NaOH → NaCl + H2O

In a titration between hydrochloric acid (HCl) and sodium hydroxide (NaOH), the reaction involved is the neutralization reaction between an acid and a base. The balanced equation for this reaction is:

HCl + NaOH → NaCl + H2O

In this reaction, one mole of HCl reacts with one mole of NaOH to form one mole of NaCl (sodium chloride) and one mole of water.

During the titration process, the reaction occurs gradually as the base is added to the acid solution.

The first end point of the titration is reached when the moles of HCl and NaOH are stoichiometrically equivalent, meaning they react in a 1:1 ratio. At this point, all the HCl has been neutralized by the NaOH, and no excess of either reagent remains.

However, if the titration is continued beyond the first end point, the reaction between HCl and NaOH can still occur, albeit in a different ratio.

The second end point refers to the point where the moles of NaOH added exceed the stoichiometrically required amount to neutralize the HCl completely. As a result, any excess NaOH added after the second end point reacts with the excess HCl in a 1:1 ratio.

Therefore, the median reaction at the second end point in the HCl and NaOH titration is:

HCl + NaOH → NaCl + H2O

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Related Questions

A 75.0- mL
volume of 0.200 M
NH3
( Kb=1.8×10−5
) is titrated with 0.500 M
HNO3
. Calculate the pH
after the addition of 17.0 mL
of HNO3
.

Answers

Answer:

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Explanation:

i am going to solve this problem by using the ICE table method which is an easy method to determine the pH of a weak base with the given data of the problem.Given:Initial volume of NH3 solution (Vi) = 75.0 mLInitial concentration of NH3 solution (Ci) = 0.200 MInitial moles of NH3 solution (Ni) = Ci x Vi = 0.200 M x 75.0 mL = 0.0150 molesKb = 1.8 x 10^-5Moles of HNO3 added (n) = 0.500 M x 17.0 mL = 0.00850 molesVolume of NH3 solution after the addition of HNO3 (Vf) = 75.0 mL + 17.0 mL = 92.0 mLConcentration of NH3 solution after the addition of HNO3 (Cf) = Ni / Vf = 0.0150 moles / 92.0 mL = 0.163 MTo find the pH after the addition of 17.0 mL of HNO3, we need to use the ICE table method.ICE table method:Initial: NH3 + H2O ⇌ NH4+ + OH-Change: -x 0 +x +xEquilibrium: 0.0150 - x 0 x xKb = [NH4+][OH-] / [NH3]1.8 x 10^-5 = x^2 / 0.163Solving for x, x = 0.00171 M[OH-] = 0.00171 M[OH-] = Kw / [H3O+] = 1.0 x 10^-14 / [H3O+][H3O+] = 5.85 x 10^-12pH = -log[H3O+]pH = -log(5.85 x 10^-12)pH = 11.23Therefore, the pH after the addition of 17.0 mL of HNO3 is approximately 11.23.

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10. When dissolved in water, most Group 1 metal salts can be described as
strong electrolytes.
strong acids.
weak electrolytes.
A
B
C
D
non-electrolytes.
(1)

Answers

When dissolved in water, most Group 1 metal salts can be described as strong electrolytes.

When Group 1 metal salts are dissolved in water, they can be described as strong electrolytes. This is because Group 1 metals, such as lithium (Li), sodium (Na), potassium (K), and so on, readily lose their outermost valence electron to form positive ions (cations). These cations then dissociate completely in water, separating from the anions to which they were originally bonded.

The dissociation of Group 1 metal salts in water results in the formation of positively charged metal ions and negatively charged non-metal ions (anions). These ions are free to move and conduct electric current, making the solution a good conductor of electricity. The complete dissociation of Group 1 metal salts in water and the presence of freely moving ions make them strong electrolytes.

Strong electrolytes are substances that ionize completely or almost completely in solution, producing a high concentration of ions. This is in contrast to weak electrolytes, which only partially ionize and produce a lower concentration of ions.

In summary, when Group 1 metal salts are dissolved in water, they form strong electrolytes due to their ability to dissociate completely into ions, leading to a high concentration of freely moving ions in the solution, thus enabling efficient electrical conductivity.

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