Answer: -
1.8
End point passed.
Explanation: -
Volume of HI solution = 47.0 mL = 0.047 L
Strength of HI solution = 0.47 M
Since HI is a strong acid, all of HI will dissociate to give H +.
[H+ ] =0.47 M x 0.047 L
= 0.02209 mol
Volume of KOH = 25.0 mL = 0.025 L
Strength of KOH = 0.25 M
Since KOH is a strong base, all of KOH will dissociate to give OH-.
[OH-] = 0.25 M x 0.025L
= 0.00625 mol
Since [H+] and [OH-] react to form water,
[H+] unreacted = 0.02209 – 0.00625 = 0.01584 mol
Using the formula
pH = - log [H+]
= - log 0.01584
= 1.8
As the strong acid HI is being titrated by strong base KOH, the pH at the end point should be 7.
The pH has already crossed that. Thus the titration end point has already passed
After titration, there are more moles of HI than KOH, implying excess HI (acid) is present. The remaining acid concentration is 0.2 M and consequently, the final pH of the solution is approximately 0.70.
In the case of the titration of a 47.0 mL of 0.47 M HI solution with 25.0 mL of 0.25 M KOH, we first need to understand that HI is a strong acid and KOH is a strong base. When we titrate a strong acid with a strong base, the equivalence point occurs at a pH of 7.0.
First, we calculate the moles of the acid and the base: moles of HI = 0.47 mol/L * 0.047 L = 0.02209 mol, and moles of KOH = 0.25 mol/L * 0.025 L = 0.00625 mol. Since there are more moles of HI than KOH, we will have extra HI left after the titration. Hence, it is a strong acid-strong base titration before the equivalence point i.e. when we have excess acid.
The remaining acid concentration is (0.02209 mol - 0.00625 mol) / (0.047 L + 0.025 L) = 0.2 M and pH of a strong acid is basically the negative logarithm of the acid's concentration. Therefore, the pH is -log[H+] = -log(0.2) = approx. 0.70.
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Answer:
Explanation:
Sn + 2HBr = SnBr₂ + H₂
Here HBr is an acid but Sn is not a base . It is a metal . So it is not an acid - base reaction .
HCl + KOH = KCl + H₂O
HCl is an acid and KOH is a base so it is an acid base reaction.
2AlCl₃ + 3Ca(OH)₂ = 2Al(OH)₃ + 3CaCl₂
It is an acid base reaction . It is so because aluminium hydroxide is a lewis acid and calcium hydroxide is a base . So it is an acid base reaction .
2C₂H₆ +7O₂ = 4CO₂ + 6H₂O
It is not an acid base reaction . It is actually an example of oxidation reaction in which ethane burns in oxygen to give carbon dioxide and water.
Explanation:
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Answer:
69.3%
Explanation:
The question should read as follows:
A weak acid, HA, has a pKa of 4.357. If a solution of this acid has a pH of 4.005, what percentage of the acid is not ionized? Assume all H⁺ in the solution came from the ionization of HA.
The Henderson-Hasselbalch equation relates the pKa and pH of a solution to the ratio of ionized (A⁻) and unionized (HA) forms of a weak acid:
pH = pKa + log ([A⁻]/[HA])
Substituting and solving for [A⁻]/[HA]:
4.005 = 4.3574 = log([A⁻]/[HA])
-0.3524 = log([A⁻]/[HA])
[A⁻]/[HA] = 0.444/1
The percentage of acid that is not ionized (i.e. the percentage of acid in the HA form) is calculated:
[HA]/([A⁻] + [HA]) x 100% = 1/(1+0.444) x 100% = 69.3%
B) 1-chloro-5-ethyl-4-methyl-3-heptene x
C) 1-chloro-3-pentenyl-2-pentene x
D) cis-7-chloro-3-ethyl-4-methyl-3-heptene
E) trans-7-chloro-3-ethyl-4-methyl-3-heptene
The question is incomplete. The complete question is attached below.
Answer : The correct option is, (A) 7-chloro-3-ethyl-4-methyl-3-heptene
Explanation :
The rules for naming of alkene are :
First select the longest possible carbon chain.
The longest possible carbon chain should include the carbons of double bonds.
The naming of alkene by adding the suffix -ene.
The numbering is done in such a way that first carbon of double bond gets the lowest number.
The carbon atoms of the double bond get the preference over the other substituents present in the parent chain.
If two or more similar alkyl groups are present in a compound, the words di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.
The given compound name will be, 7-chloro-3-ethyl-4-methyl-3-heptene.
The structure of given compound is shown below.
The structure given can be named as trans-7-chloro-3-ethyl-4-methyl-3-heptene.
The structure given can be named as trans-7-chloro-3-ethyl-4-methyl-3-heptene. The name is determined by identifying the longest carbon chain, the substituents attached to it, and their positions. In this case, the longest carbon chain has 7 carbons, with a chlorine atom attached at position 7. There is an ethyl group at position 3 and a methyl group at position 4. The presence of double bonds is indicated by the -ene ending.
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