A buffer containing 0.2000M of acid, HA, and 0.1500M of its...

A buffer containing 0.2000$M$ of acid, HA, and 0.1500$M$ of its conjugate base, $A^{-},$ has a pH of $3.35 .$ What is the pH after 0.0015 mol of $\mathrm{NaOH}$ is added to 0.5000 $\mathrm{L}$ of this solution?

Key Concepts

Stoichiometry in Buffer Reactions

In buffer systems, the addition of a strong acid or base results in a stoichiometric reaction where the strong base reacts with the weak acid (or vice-versa) to form more of the conjugate base. This reaction alters the concentrations of the buffer components without completely overwhelming the buffer capacity, thereby enabling the calculation of the new pH by adjusting the ratio in the Henderson-Hasselbalch equation.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation provides a straightforward way to calculate the pH of a buffer solution using the relationship: pH = pKa + log([Conjugate Base]/[Acid]). This equation is essential in understanding how changes in the ratio of the buffer components affect the pH, especially when the composition of the buffer is altered by adding an acid or base.

Buffer Solutions

Buffers are mixtures of a weak acid and its conjugate base (or a weak base and its conjugate acid) that resist large pH changes upon the addition of small amounts of strong acid or base. The equilibrium between the acid and its conjugate base allows the solution to neutralize added H+ or OH-, thereby maintaining a relatively stable pH.

Transcript

00:01 Some weak acid, ha, partially dissociates into a -minus and h -3 -0 -plus, and this forms a buffer.

00:09 If this buffer has a ph of 3 .35, where the concentration of h -a is 0 .2 molar, and the concentration of a -minus 0 .15 molar, we can calculate the p .k .a.

00:29 Of h .a.

00:30 Using the henderson -hasselbach equation.

00:45 Rearranging this, we get p .k .a.

00:48 Equals ph minus the log of the proper component concentration ratio, which is the concentration of a minus over the concentration of h .a.

01:01 So we can plug our numbers in here to get 3 .35 minus the log of 0 .3 .3 .3 .3 .5 minus the log of 0 .15 over 0 .2.

01:19 This is 3 .35 minus negative 0 .1249, which is 3 .4749.

01:36 I'd say we add 0 .0015 moles of n -aoh to 0 .5 liters of this buffer solution.

01:50 The concentration of nah since molarity is moles over liters.

01:56 So we can divide 0 .015 moles by 0 .5 liters and get 0 .003 molar nahoh.

02:11 Some weak.

02:12 The weak acid from the buffer will react with the hydroxide ions from the added nah to form water and conjugate base.

02:28 Since hydroxide is a strong base, it will fully react with the weak acid, which is why this is a single forward arrow.

02:39 So now we can set up an ice table.

02:44 Initially we have 0 .2 -0 -00 molar hae, 0 .003 molar oh h -minus from the added and a .o .h and 0 .150 molar a minus.

03:03 Liquids aren't included in these calculations, so we can just ignore water here...

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