What volumes of 0.200M HCOOH and 2.00M NaOH would make 500 . mL of a buffer with the same pH as

step 1 Question 105 is a buffer solution calculation. For buffer solution calculations, we typically us

What volumes of 0.200M HCOOH and 2.00M NaOH would make 500 ....

Key Concepts

pKa and Its Role in Buffer pH

The pKa of a weak acid is the pH at which the acid is half dissociated. It is a fundamental property in buffer chemistry since the effective buffering range of a weak acid is usually within about one pH unit of its pKa. Matching or adjusting the acid–base ratio in light of the acid's pKa allows for tailoring the buffer to achieve a specific pH.

Henderson–Hasselbalch Equation

The Henderson–Hasselbalch equation, pH = pKa + log([conjugate base]/[acid]), is a key tool that relates the pH of a buffer solution to the ratio of the concentrations of the conjugate base and the weak acid. This equation is used to design buffers with a desired pH by setting the appropriate acid-to-base ratio.

Buffer Solutions

A buffer solution is one that resists changes in pH when small amounts of acid or base are added. Buffers are typically composed of a weak acid and its conjugate base (or a weak base and its conjugate acid) in roughly comparable amounts that work together to stabilize the pH of the solution.

Acid–Base Neutralization and Stoichiometry

In preparing a buffer, an acid reacts partially with a strong base through a neutralization reaction to form its conjugate base. The stoichiometry of this reaction, which determines how many moles of acid are converted and how many remain, is critical to establishing the correct ratio of acid to conjugate base in the final buffer solution.

Transcript

00:01 Question 105 is a buffer solution calculation.

00:06 For buffer solution calculations, we typically use the henderson -hasselbalch equation.

00:12 However, this question is a little different in that it gives you concentrations associated with one buffer for which you need to calculate the ph, and then knowing the ph of that one buffer, calculate the concentrations of the other species needed for a different buffer.

00:35 So we know the concentrations exactly for the second buffer.

00:40 So let's calculate the concentrations for the second or let's calculate the ph that is created from the concentrations of the second buffer and then we can use that ph of the second buffer to calculate the concentrations required to achieve that same ph of the first buffer.

00:59 So the second buffer is comprised of benzoic acid and some sodium hydroxide.

01:08 So we need to know when using the henderson hasselbalch equation, the k .a.

01:13 Value for benzoic acid.

01:16 The henderson -hasselbalch equation is ph equals p -ka, which is the negative log of the ka for the weak acid in the buffer, which is benzoic acid, plus the log of the moles of base over the moles of acid.

01:32 The moles of base will be defined by the moles of strong base added.

01:36 Every time we add a mole of sodium hydroxide to a solution that contains benzoic acid, one mole of benzoic acid is consumed, and one mole of benzate, its conjugate base, is formed.

01:50 So because we added 25 milliliters of two molar sodium hydroxide, then if we take 25 milliliters, 0 .025 liters, multiply by the molarity of the solubyxide, sodium hydroxide.

02:04 This will give us the moles of sodium hydroxide added, which will now be the moles of benzoyate, the weak base formed.

02:11 The moles of weak acid that will be present will be the volume of the benzoic acid solution, 475 milliliters or 0 .475 liters, multiplied by the concentration of the benzoic acid.

02:27 This is the original moles of benzoic acid before adding the sodium hydroxide, but when we add the sodium hydroxide, the moles of benzoyate formed will be equal to the moles of benzoic acid that was decreased.

02:43 So now we have ph equal to pca plus the log of the moles of the weak base that are present, divided by the moles of the weak acid that are present after adding the sodium hydroxide.

02:56 And we get a ph of 4 .246.

03:00 So now to determine the concentrations that are required for the first buffer in order to achieve the ph of that second buffer which is 4 .246 we need to take the negative log of the weak acid that is present in this first buffer its formic acid with a k a value of 1 .8 times 10 to the negative 4 so its pka value would be the negative log of 1 .8 times 10 to the negative 4 so ph equals pkk plus the log of the moles of base over the moles of acid.

03:38 Well, this is all we know.

03:42 We have the molarity of the sodium hydroxide that we are going to add in order to create our formic acid formate buffer.

03:55 For every mole of sodium hydroxide that we add, we will convert one mole of formic acid into formate.

04:03 We don't know the volume of sodium hydroxide that we're going to add, so we will define that as x...

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