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seraphim [82]
3 years ago
5

A 25.0 ml sample of an unknown monoprotic acid was titrated with 0.12 m naoh. the student added 31.6 ml of naoh and went past th

e equivalence point. which procedure could be performed next to more accurately determine the concentration of the unknown acid?
Chemistry
1 answer:
Inessa05 [86]3 years ago
7 0
The procedure, which can be used to determine more accurately the concentration of the unknown acid is TO BACK-TITRATE WITH ADDITIONAL HYDROCHLORIC ACID TO NEUTRALIZE THE ADDITIONAL SODIUM HYDROXIDE THAT WAS ADDED.
Monoprotic acids are acids that can donate only one proton per each molecule and they have only one equivalence point. Examples of monoprotic acids are HCI, HNO3 and CH3COOH.
The back titration method is typically used when one needs to determine the concentration of an analyte provided there is a known molar concentration of excess reactants. 
From the information given in the question above, we are told that excess NaOH was added. To correct this mistake, the right thing to do is to use additional HCl to carry out back titration, taking note of the quantity of acid that will be needed to neutralize the excess NaOH.
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The solubility of agcl(s) in water at 25 ∘c is 1.33×10−5mol/l and its δh∘ of solution is 65.7 kj/mol. what is the solubility at
slamgirl [31]
According to this equation:

AgCl(s) ↔ Ag+(aq)  + Cl-(aq)

so K1 = [Ag+][Cl-]

when [Ag+] = [Cl-]  we can assume both = X 

and when we have X the solubility = 1.33 x 10^-5 mol / L

by substitution:

∴ K1 = X^2

       = (1.33 x 10^-5)^2

       = 1.77 x 10^-10

by using vant's Hoff equation:

ln(K2/K1) = (ΔH/R)*(1/T2-1/T1)

when ΔH = 65700 J / mol

R = 8.314 

T1 = 25+273 = 298 K

T2 = 47.7 +273 =320.7

by substitution:

∴㏑(K2/1.77 x 10^-10) = (65700/8.314) * ( 1/320.7 - 1/ 298)

by solving for K2 

∴K2 = 2.7 x 10^-11

and when K2 = X^2

∴ the solubility X = √(2.7 x 10^-11)

                             = 5.2 x 10^-6 mol/L  
8 0
3 years ago
A student wants to make an electromagnet. Which items are needed?
bazaltina [42]

Answer:

copper wire, metal nail, battery

Explanation:

To create an electromagnet, the material to be magnetized (such as a metal nail) is placed inside a solenoid, the ends of the coils are connected to a circuit that includes a battery.

Current is allowed to flow for a few seconds and then cut off. The polarity of the magnet depends on the direction of flow of current in the solenoid.

8 0
3 years ago
When 18.0 mL of a 8.43×10-4 M cobalt(II) fluoride solution is combined with 22.0 mL of a 9.72×10-4 M sodium hydroxide solution d
Eva8 [605]

Answer:

Q = 1.08x10⁻¹⁰

Yes, precipitate is formed.

Explanation:

The reaction of CoF₂ with NaOH is:

CoF₂(aq) + 2 NaOH(aq) ⇄ Co(OH)₂(s) + 2 NaF(aq).

The solubility product of the precipitate produced, Co(OH)₂, is:

Co(OH)₂(s) ⇄ Co²⁺(aq) + 2OH⁻(aq)

And Ksp is:

Ksp = 3x10⁻¹⁶= [Co²⁺][OH⁻]²

Molar concentration of both ions is:

[Co²⁺] = 0.018Lₓ (8.43x10⁻⁴mol / L) / (0.018 + 0.022)L = <em>3.79x10⁻⁴M</em>

[OH⁻] = 0.022Lₓ (9.72x10⁻⁴mol / L) / (0.018 + 0.022)L = <em>5.35x10⁻⁴M</em>

Reaction quotient under these concentrations is:

Q = [3.79x10⁻⁴M] [5.35x10⁻⁴M]²

<em>Q = 1.08x10⁻¹⁰</em>

As Q > Ksp, <em>the equilibrium will shift to the left producing Co(OH)₂(s) </em>the precipitate

8 0
4 years ago
Solid silicon and solid magnesium chloride form when silicon tetrachloride gas reacts with magnesium metal. Write a word equa- t
antiseptic1488 [7]
SiCl4 + 2Mg ---> 2Si + 2MgCl2
8 0
3 years ago
The IUPAC rules permit the use of common names for a number of familiar phenols and aryl ethers. These common names are listed h
fgiga [73]

Explanation:

a. Vanillin(4-hydroxy-3-methoxybenzaldehyde):

In its structure hydroxl group will be present on para position of the benzaldehyde ring and methoxy group on meta position.

b. Thymol (2-isopropyl-5-methylphenol):

In its structure isopropyl group will be present on ortho position of the phenol ring and methyl group on meta position.

c. Carvacrol (5-isopropyl-2-methylphenol):

In its structure isopropyl group will be present opposite to methyl group which is present ortho position in a phenol ring.

d. Eugenol (4-allyl-2-methoxyphenol):.

In its structure allyl group will be present on para position of the phenol ring and methoxy group on ortho position.

e. Gallic acid (3,4,5-trihydroxybenzoic acid):

In its structure hydroxyl group will be present on both meta positions and on para position of the benzoic acid ring.

f. Salicyl alcohol (o-hydroxybenzyl alcohol):

In its structure, -CH_2OH group is linkedto benzene ring and in respect to that hydroxyl group is present at ortho position of the ring.

6 0
3 years ago
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