what volume of 6.00m naoh would be required to increase the ph to 4.93?

April 03, 2022 Post a Comment

Buffer Solutions

A buffer solution is i in which the pH of the solution is "resistant" to minor additions of either a strong acid or stiff base. Buffers usually consist of a weak acrid and its conjugate base, in relatively equal and "big" quantities. Calculations are based on the equation for the ionization of the weak acid in water forming the hydronium ion and the conjugate base of the acid. "HA" represents whatsoever weak acid and "A^-" represents the conjugate base.

HA(aq) + H₂O(fifty) --> H_threeO⁺(aq) + A^-(aq)

G_a = [H ₃ O ⁺ ][A ^- ]

[HA]

A buffer system tin be made by mixing a soluble compound that contains the cohabit base with a solution of the acid such as sodium acetate with acetic acid or ammonia with ammonium chloride. The above equation for Thou_a can be rearranged to solve for the hydronium ion concentration. By knowing the K_a of the acrid, the corporeality of acrid, and the amount of conjugate base, the pH of the buffer system can exist calculated.

[H_iiiO⁺] = K_a [HA] [A^-] pH = -log[H₃O⁺]

Calculation of the pH of a Buffer Solution
Calculation of the pH of a Buffer Solution afterwards Improver of a Pocket-sized Amount of Strong Acid
Calculation of the pH of a Buffer Solution after Add-on of a Small Amount of Strong Base
Calculation of the Buffer Capacity

Adding of the pH of a Buffer Solution

In lodge to summate the pH of the buffer solution you need to know the amount of acid and the amount of the conjugate base combined to make the solution. These amounts should be either in moles or in molarities. The Yard_a of the acid also needs to exist known.

Example: A buffer solution was made by dissolving 10.0 grams of sodium acetate in 200.0 mL of i.00 M acetic acid. Assuming the alter in book when the sodium acetate is not significant, estimate the pH of the acetic acid/sodium acetate buffer solution. The One thousand_a for acerb acid is 1.seven x 10^-5.

First, write the equation for the ionization of acerb acid and the G_a expression. Rearrange the expression to solve for the hydronium ion concentration.

CH₃COOH(aq) + H₂O(l) --> H₃O⁺(aq) + CH₃COO^-(aq)

[H₃O⁺] = K_a [CH ₃ COOH]

[CH_iiiCOO^-]

Second, determine the number of moles of acrid and of the cohabit base of operations.

(i.00 Thou CH₃COOH)(200.0 mL)(1 L/1000 mL) = 0.200 mol CH₃COOH

(10.0 chiliad NaCH_iiiCOO)(1 mol/82.03 g) = 0.122 mol NaCH_threeCOO

Substitute these values, along with the K_a value, into the above equation and solve for the hydronium ion concentration. Convert the hydronium ion concentration into pH.

[H₃O⁺] = (1.7 x 10^-5)(0.200/0.122) = two.79 x ten^-5
pH = four.56

Example: Calculate the ratio of ammonium chloride to ammonia that is required to brand a buffer solution with a pH of 9.00. The Ka for ammonium ion is v.6 ten 10^-10.

First, write the equation for the ionization of the ammonium ion in water and the corresponding Ka expression. Rearrange the equation to solve for the hydronium ion concentration.

NH₄ ⁺(aq) + H₂O(l) --> H_iiiO⁺(aq) + NH_iii(aq)

K_a = [H _iii O ⁺ ][NH ₃ ]

[NH₄ ⁺]

[H_iiiO⁺] = K_a [NH ₄ ⁺ ]

[NH₃]

Second, convert the pH dorsum into the hydronium ion concentration and then substitute information technology into the above equation along with the K_a. Solve for the ratio of ammonium ion to ammonia.

[H_threeO⁺] = 1 ten 10^-ix M

1 x 10^-ix = 5.half dozen 10 x^-10(NH_four ⁺/NH₃)
(NH₄ ⁺/NH_three) = 1.786/i

A ratio of 1.768 moles of ammonium ion for every 1 mole of ammonia or ane.768 Thousand ammonium ion to 1 1000 ammonia.

Acme

Adding of the pH of a Buffer Solution subsequently Improver of a Minor Amount of Acid

When a strong acrid (H₃O⁺) is added to a buffer solution the conjugate base present in the buffer consumes the hydronium ion converting it into h2o and the weak acid of the conjugate base.

A^-(aq) + H₃O⁺(aq) --> H_iiO(fifty) + HA(aq)

This results in a decrease in the amount of conjugate base nowadays and an increase in the corporeality of the weak acrid. The pH of the buffer solution decreases by a very small amount because of this ( a lot less than if the buffer system was not present). An "ICE" chart is useful in determining the pH of the system afterward a potent acid has been added.

Example: fifty.0 mL of 0.100 K HCl was added to a buffer consisting of 0.025 moles of sodium acetate and 0.030 moles of acerb acid. What is the pH of the buffer after the addition of the acrid? Ka of acerb acid is i.7 x 10^-5.

Get-go, write the equation for the ionization of acerb acid in h2o and the related M_a expression rearranged to solve for the hydronium ion concentration.

CH₃COOH(aq) + H_iiO(l) --> H_threeO⁺(aq) + CH_iiiCOO^-(aq)

[H₃O⁺] = Grand_a [CH _iii COOH]

[CH₃COO^-]

Second, brand an "ICE" chart. Let "x" stand for the hydronium ion concentration one time equilibrium has been re-established. We will assume that all of the added acid is consumed.

	CH_iiiCOOH(aq)	H_iiiO⁺(aq)	CH₃COO^-(aq)
Initial Amount	0.030 moles	(0.0500 Fifty)(0.100 Grand) = 0.0050 moles	0.025 moles
Modify in Amount	+ 0.005 moles	-0.005 moles	- 0.005 moles
Equilibrium Amount	0.035 moles	10	0.020 moles

Substitute into the K_a expression and solve for the hydronium ion concentration. Convert the answer into pH.

[H₃O⁺] = (ane.7 10 x^-5)(0.035/0.020) = 2.975 x 10^-v
pH = 4.53

Pinnacle

Calculation of the pH of a Buffer Solution after Addition of a Pocket-size Amount of Strong Base

When a strong base (OH^-) is added to a buffer solution, the hydroxide ions are consumed by the weak acid forming water and the weaker conjugate base of the acid. The amount of the weak acid decreases while the amount of the conjugate base increases. This prevents the pH of the solution from significantly rising, which it would if the buffer system was not present.

OH^-(aq) + HA(aq) --> H_twoO(fifty) + A^-(aq)

The process for finding the pH of the mixture after a strong base has been added is similar to the addition of a stiff acid shown in the previous department.

Case: Calculate the pH of a buffer solution that initially consists of 0.0400 moles of ammonia and 0.0250 moles of ammonium ion, after 20.0 mL of 0.75 M NaOH has been added to the buffer. Ka for ammonium ion is 5.6 x 10^-10.

Showtime, write the equation for the ionization of the ammonium ion and the related Thou_a expression solved for the hydronium ion concentration.

NH₄ ⁺(aq) + H_iiO(l) --> H₃O⁺(aq) + NH₃(aq)

[H_threeO⁺] = 1000_a [NH ₄ ⁺ ]

[NH_three]

Second, make an "Ice" chart. Allow "10" be the concentration of the hydronium ion at equilibrium. The alter in the amount of the ammonium ion volition be equal to the amount of potent base added (075 Thousand x 0.0200 Fifty = 0.0015 mol).

	NH_four ⁺(aq)	H₃O⁺(aq)	NH_iii(aq)
Initial Amount	0.0250 moles	* not needed	0.0400 moles
Change in Corporeality	- 0.0015 moles	* not needed	+ 0.0015 moles
Equilibrium Amont	0.0235 moles	x	0.0415 moles

Third, substitute into the K_a expression and solve for the hydronium ion concentration. Convert the answer into pH.

[H_threeO⁺] = (v.half-dozen x 10^-10)(0.0235/0.0415) = 3.17 x 10^-10
pH = ix.fifty

Superlative

Calculation of the Buffer Chapters

The buffer capactity refers to the maximum amount of either strong acid or strong base that can be added before a significant modify in the pH will occur. This is only a matter of stoichiometry. The maximum amount of potent acid that tin be added is equal to the corporeality of conjugate base present in the buffer. The maximum amount of base of operations that can be added is equal to the amount of weak acid present in the buffer.

Example: What is the maximum amount of acrid that can be added to a buffer made by the mixing of 0.35 moles of sodium hydrogen carbonate with 0.l moles of sodium carbonate? How much base of operations tin can be added earlier the pH will brainstorm to evidence a pregnant change?

Get-go, write the equation for the ionization of the weak acid, in this example of hydrogen carbonate. Although this footstep is not truly necessary to solve the problem, it is helpful in identifying the weak acid and its cohabit base of operations.

HCO₃ ^-(aq) + H₂O(l) --> H₃O⁺(aq) + CO_iii ^2-(aq)

Second, added potent acid will react with the conjugate base, CO₃ ^ii-. Therefore, the maximum corporeality of acrid that can be added will exist equal to the amount of CO₃ ^2-, 0.50 moles.
Third, added stiff base will react with the weak acrid, HCO₃ ^-. Therefore, the maximum amount of base that tin can be added will be equal to the amount of HCO₃ ^-, 0.35 moles.

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