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3 years ago

Classify each solvent as protic or aprotic. a. (CH3)2CHOH c. CH2Cl2 e. N(CH3)3 b. CH3NO2 d. NH3 f. HCONH2

Chemistry

1 answer:

xxTIMURxx [149]3 years ago

3 0

Answer:

The protic solvents are (CH3)2CHOH, NH3, and HCONH2

The aprotic solvent are CH3NO2, CH2Cl2, and N(CH3)3

Explanation:

A protic solvent is a solvent that has a hydrogen atom bound to an oxygen (O), a nitrogen (N), or fluoride (F). Generally, any solvent that contains a labile H⁺ is called a protic solvent.

a. (CH3)2CHOH

(CH3)2CHOH is a protic solvent because one of the hydrogens is bound to oxygen (O)

b. CH3NO2

CH3NO2 is an aprotic (i.e not a protic solvent) solvent because all the hydrogens are bound to carbon.

c. CH2Cl2

CH2Cl2 is an aprotic solvent because all the hydrogens are bound to carbon.

d. NH3 is a protic solvent because all the hydrogens are bound to Nitrogen (N)

e. N(CH3)3 is an aprotic solvent because all the hydrogens are bound to carbon

f. HCONH2 is a protic solvent because two of the hydrogens are bound to Nitrogen (N)

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3 years ago

what volume of a 0.138 m potassium hydroxide solution is required to neutralize 26.0 ml of a 0.205 m nitric acid solution?

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A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is ( $M1V1 = M2V2$ ).

Utilize the titration method of $M1V1 = M2V2$ in view that we're given the concentrations of every compound and the quantity of $KOH$ . Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.

M1 = initial mass
V1= initial volume
M2 = final mass
V2= final volume

$(M1V1 = M2V2)$
(0.138)(V1) = (0.205)x(26.0)
V2=(0.205)x(26.0)\ 0.138
V2 = 47.10 M/L
The final value of Volume needed for neutralization of nitric acid solution is V2 = 47.10 M/L

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1 year ago

For the titration of 25.00 mL of 0.150 M HCl with 0.250 M NaOH, calculate:

Leviafan [203]

1) Chemical reaction

HCl        +       NaOH      --->      NaCl + H2O

25.0 ml
0.150 M            0.250M

2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution

0.001875 mol HCl => 0.001875 mol H(+)

Volume = Volume of HCl solution + Volumen of NaOH solution added

Volume of HCl solution = 0.0250 l

Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l

Total volume = 0.0250 l + 0.0075 l = 0.0325 l

[H+] = 0.001875 mol / 0.0325 l = 0.05769 M

pH = - log [H+] = - log (0.05769) = 1.23

Answer: 1.23

3) Equivalence point

0.02500 l * 0.150 M = 0.250M * V

=> V = 0.02500 * 0.150 / 0.250 = 0.015 l

4) 1.00 ml NaOH added beyond the equivalence point

1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess

0.00025 mol NaOH = 0.00025 mol OH-

Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l

[OH-] = 0.00025 mol / 0.041 l = 0.00610 M

pOH = - log (0.00610) = 2.21

pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76

Answer: 11.76

6 0

3 years ago