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Find the Ph of a 0.250 M Solution of Nac2h3o2

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Buffer Solutions

A buffer solution is one in which the pH of the solution is "resistant" to small additions of either a strong acid or strong base.  Buffers usually consist of a weak acid and its conjugate base, in relatively equal and "large" 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 any weak acid and "A-" represents the conjugate base.

HA(aq) + H2O(l) --> H3O+(aq) + A-(aq)

Ka = [H 3 O + ][A - ]

[HA]

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

[H3O+] = Ka [HA]

[A-]

pH = -log[H3O+]
  • Calculation of the pH of a Buffer Solution
  • Calculation of the pH of a Buffer Solution after Addition of a Small Amount of Strong Acid
  • Calculation of the pH of a Buffer Solution after Addition of a Small Amount of Strong Base
  • Calculation of the Buffer Capacity

Calculation of the pH of a Buffer Solution

In order to calculate 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 Ka of the acid also needs to be known.

Example:  A buffer solution was made by dissolving 10.0 grams of sodium acetate in 200.0 mL of  1.00 M acetic acid.  Assuming the change in volume when the sodium acetate is not significant, estimate the pH of the acetic acid/sodium acetate buffer solution.  The Ka for acetic acid is 1.7 x 10-5.

  • First, write the equation for the ionization of acetic acid and the Ka expression.  Rearrange the expression to solve for the hydronium ion concentration.
CH3COOH(aq) + H2O(l) --> H3O+(aq) + CH3COO-(aq)

[H3O+] = Ka [CH 3 COOH]

[CH3COO-]

  • Second, determine the number of moles of acid and of the conjugate base.
(1.00 M CH3COOH)(200.0 mL)(1 L/1000 mL) = 0.200 mol CH3COOH

(10.0 g NaCH3COO)(1 mol/82.03 g) = 0.122 mol NaCH3COO

  • Substitute these values, along with the Ka 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) = 2.79 x 10-5
pH = 4.56

Example:  Calculate the ratio of ammonium chloride to ammonia that is required to make a buffer solution with a pH of 9.00.  The Ka for ammonium ion is 5.6 x 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) --> H3O+(aq) + NH3(aq)

Ka = [H 3 O + ][NH 3 ]

[NH4 +]

[H3O+] = Ka [NH 4 + ]

[NH3]

  • Second, convert the pH back into the hydronium ion concentration and then substitute it into the above equation along with the Ka.  Solve for the ratio of ammonium ion to ammonia.
[H3O+] = 1 x 10-9 M

1 x 10-9 = 5.6 x 10-10(NH4 +/NH3)
(NH4 +/NH3) = 1.786/1

A ratio of 1.768 moles of ammonium ion for every 1 mole of ammonia or 1.768 M ammonium ion to 1 M ammonia.

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Calculation of the pH of a Buffer Solution after Addition of a Small Amount of Acid

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

A-(aq) + H3O+(aq) --> H2O(l) + HA(aq)

This results in a decrease in the amount of conjugate base present and an increase in the amount of the weak acid.  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 after a strong acid has been added.

Example: 50.0 mL of 0.100 M HCl was added to a buffer consisting of 0.025 moles of sodium acetate and 0.030 moles of acetic acid.  What is the pH of the buffer after the addition of the acid?  Ka of acetic acid is 1.7 x 10-5.

  • First, write the equation for the ionization of acetic acid in water and the related Ka expression rearranged to solve for the hydronium ion concentration.
CH3COOH(aq) + H2O(l) --> H3O+(aq) + CH3COO-(aq)

[H3O+] = Ka [CH 3 COOH]

[CH3COO-]

  • Second, make an "ICE" chart.  Let "x" represent the hydronium ion concentration once equilibrium has been re-established.  We will assume that all of the added acid is consumed.
CH3COOH(aq)
H3O+(aq)
CH3COO-(aq)
Initial Amount
0.030 moles
(0.0500 L)(0.100 M) = 0.0050 moles
0.025 moles
Change in Amount
+ 0.005 moles
-0.005 moles
- 0.005 moles
Equilibrium Amount
0.035 moles
x
0.020 moles
  • Substitute into the Ka expression and solve for the hydronium ion concentration.  Convert the answer into pH.
[H3O+] = (1.7 x 10-5)(0.035/0.020) = 2.975 x 10-5
pH = 4.53

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Calculation of the pH of a Buffer Solution after Addition of a Small 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) --> H2O(l) + 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 strong acid shown in the previous section.

Example: 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.

  • First, write the equation for the ionization of the ammonium ion and the related Ka expression solved for the hydronium ion concentration.
NH4 +(aq) + H2O(l) --> H3O+(aq) + NH3(aq)

[H3O+] = Ka [NH 4 + ]

[NH3]

  • Second, make an "ICE" chart.  Let "x" be the concentration of the hydronium ion at equilibrium.  The change in the amount of the ammonium ion will be equal to the amount of strong base added (075 M x 0.0200 L = 0.0015 mol).
NH4 +(aq)
H3O+(aq)
NH3(aq)
Initial Amount
0.0250 moles
* not needed
0.0400 moles
Change in Amount
- 0.0015 moles
* not needed
+ 0.0015 moles
Equilibrium Amont
0.0235 moles
x
0.0415 moles
  • Third, substitute into the Ka expression and solve for the hydronium ion concentration.  Convert the answer into pH.
[H3O+] = (5.6 x 10-10)(0.0235/0.0415) = 3.17 x 10-10
pH = 9.50

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Calculation of the Buffer Capacity

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

Example:  What is the maximum amount of acid that can be added to a buffer made by the mixing of 0.35 moles of sodium hydrogen carbonate with 0.50 moles of sodium carbonate?  How much base can be added before the pH will begin to show a significant change?

  • First, write the equation for the ionization of the weak acid, in this case of hydrogen carbonate.  Although this step is not truly necessary to solve the problem, it is helpful in identifying the weak acid and its conjugate base.
HCO3 -(aq) + H2O(l) --> H3O+(aq) + CO3 2-(aq)
  • Second, added strong acid will react with the conjugate base, CO3 2-.  Therefore, the maximum amount of acid that can be added will be equal to the amount of CO3 2-, 0.50 moles.
  • Third, added strong base will react with the weak acid, HCO3 -.  Therefore, the maximum amount of base that can be added will be equal to the amount of HCO3 -, 0.35 moles.
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Find The Ph Of A 0.250 M Solution Of Nac2h3o2. (The Ka Value Of Hc2h3o2 Is 1.80×10-5).

Posted by: simsaguire.blogspot.com

Source: https://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Buffers.htm

Find the Ph of a 0.250 M Solution of Nac2h3o2

Source: https://simsaguire.blogspot.com/2021/10/find-ph-of-0250-m-solution-of-nac2h3o2.html