Answer:
21.5mL of a 0.100M HCl are required
Explanation:
The sodium phenoxide reacts with HCl to produce phenol and NaCl in a 1:1 reaction.
To solve this question we need to find the moles of sodium phenoxide. These moles = Moles of HCl required to reach equivalence point and, with the concentration, we can find the needed volume as follows:
<em>Mass NaC6H5O:</em>
1.000g * 25% = 0.250g NaC6H5O
<em>Moles NaC6H5O -116.09g/mol-</em>
0.250g NaC6H5O * (1mol/116.09g) = 2.154x10⁻³ moles = Moles of HCl required
<em>Volume 0.100M HCl:</em>
2.154x10⁻³ moles HCl * (1L/0.100mol) = 0.0215L =
<h3>21.5mL of a 0.100M HCl are required</h3>
A heterozygous is represented by 1 capital letter and 1 lowercase in most instances.
Answer:
hello your question is incomplete attached below is the complete question
a) 0.12 M
b) Ka = 3.0 * 10^-7
c) Alizarin Yellow R
Explanation:
<u>A) Determine the concentration of the unknown acid</u>
The PH of the unknown acid before addition of NaOH is ; 3.72 ( weak acid )
First determine the moles of of NaOH
= molarity * volume
= 0.10 M * 30.0 ML = 0.0030 mol
at equivalence point
moles of NaOH = moles of unknown acid = 0.0030 mol
volume of unknown acid = 25.0 mL
Next calculate the concentration of the Acid ( HA )
= moles / volume
= 0.0030 mol / 25.0 mL
= 0.12 M
<u>b) Determine the Ka of the unknown acid </u>
attached below is the detailed solution
Ka = 3.0 * 10^-7
c) The indicator that would be a good choice to use in the titration of this acid with NaOH is ; Alizarin Yellow R . this is because the titration is between a strong base and a weak acid.
Answer:
Answers are in the explanation
Explanation:
Ksp of CdF₂ is:
CdF₂(s) ⇄ Cd²⁺(aq) + 2F⁻(aq)
Ksp = 6.44x10⁻³ = [Cd²⁺] [F⁻]²
When an excess of solid is present, the solution is saturated, the molarity of Cd²⁺ is X and F⁻ 2X:
6.44x10⁻³ = [X] [2X]²
6.44x10⁻³ = 4X³
X = 0.1172M
<h3>[F⁻] = 0.2344M</h3><h3 />
Ksp of LiF is:
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
Ksp = 1.84x10⁻³ = [Li⁺] [F⁻]
When an excess of solid is present, the solution is saturated, the molarity of Li⁺ and F⁻ is XX:
1.84x10⁻³ = [X] [X]
1.84x10⁻³ = X²
X = 0.0429
<h3>[F⁻] = 0.0429M</h3><h3 /><h3>The solution of CdF₂ has the higher fluoride ion concentration</h3>
Answer:
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