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The Ka value for acetic acid, CH3COOH(aq), is 1.8×10^-5. Calculate the ph of a 2.80 M…

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Question “The Ka value for acetic acid, CH3COOH(aq), is 1.8×10^-5. Calculate the ph of a 2.80 M…”

The Ka value for acetic acid, CH3COOH(aq), is 1.8×10^-5.
Calculate the ph of a 2.80 M acetic acid solution.

PH=

Calculate the ph of the resulting solution when 3.00 mL of the
2.80 M acetic acid is diluted to make a 250.0 mL solution.

PH=

Answers are not 4.6 or 3.8

Answer

Concepts and Reason

Power of hydrogen ions (pH).

The pH is the negative logarithmic function (

) of hydrogen in an aqueous solution.

Molar concentration:

The molar concentration of a solution is the amount of moles that a solute will dissolve in one liter of water. It can be expressed in mol/L (molarity) or M (molarity).

Fundamentals

Molar concentration:



Molar concentration = number of moles of solute in mol volume of solution in L



Dilution equation:



M,V, %3 М,V, 2



Acid dissociation constant



НА +Н,О —></figure><p>n0+A k HA The acid dissociation constant for</a></b> is K.<br /> </a><br /> </b></p><p> Equation to determine the pH of the solution</p><p> <b><br /> <a href=

pH logH Concentration of hydrogen ion is H*].



Henderson-Hasseslbalch Equation:

[A] pH pK,+logTHA Acid dissociation constant of HA is pK. Concentration of HA is [HA] Concentration of A is| A |.

An ICE table can be used to determine the concentrations and moles of individual products or reactants at equilibrium.

This is how an ICE table expands:


The initial quantities is denoted as I. The change in each quantity is denoted as C. The equilibrium quantities is denoted as


All quantities are expressed as moles or concentrations.

(1)

Given data:

Initial concentration of Acetic Acid,

НА

= 2,80 M


The Ka value for acetic acid, CH3COOH(aq), is 1.8x10^-5. Calculate the ph of a 2.80 M...


[но ТА] Н.о к, [НА]



Н,о — | Н,о A Н.О ICE НА 1 (М) С (М) Е (М) | 2.80-х 2.80 +x +x -х +x +x


At equilibrium: The concentration change is x.


(x)(x) к, (2.80-x) Substitute 1.8x10=(x)(x) (2.80-x)



Solve x x 7.09x103 So, H0*]-7.09x 10 M or H*7.09x 10 M




H*7.09x10 M pH-log[H Substitute pH -log (7.09x103) pH 2.15



(2)

Given:

М, 3 2.80М V 3.00mL V2 250.0mL



Using dilution equation M2 V2 M,V





Rearrange М, - М,у V, м, 2.80 Мx3.00 mL м, 250.0mL М, 3 0.0336 M



Concentration of Acetic Acid,

= 0.0336M






+| Н,о | — | Н,О| +| A ICE НА 1(М) 0.0336 С (М) Е (М) | 0.0336-х -X +x +x +x +x


At equilibrium: The concentration change is x.


(x)(x) к, (0.0336-x) Substitute (x)(x) 1.8x10 (0.0336-x)



Solvex x 7.77x10 So, HO*]-7.77x10 M or H*7.77x 10 M




H*7 pH log[H] 7.77x104 M Substitute pH=-log(7.77x10) pH 3.11



Part 1: Ans

The pH of a solution of 2.80 M Acetic Acid is



pH 2.15



Part 2

The pH of the solution is



pH 3.11



Conclusion

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