A buffer that contains 0.110M HY and 0.220M Y^- has a pH of 8.77 . What is the pH after 0.0015 m

A buffer that contains 0.110M HY and 0.220M Y^- has a pH of...

A buffer that contains 0.110$M \mathrm{HY}$ and 0.220$M \mathrm{Y}^{-}$ has a $\mathrm{pH}$ of $8.77 .$ What is the pH after 0.0015 $\mathrm{mol}$ of $\mathrm{Ba}(\mathrm{OH})_{2}$ is added to 0.350 $\mathrm{L}$ of this solution?

Key Concepts

Buffer Solution

A buffer solution is a mixture of a weak acid and its conjugate base that resists changes in pH when small amounts of acids or bases are added. This concept is central to understanding how the solution maintains a relatively constant pH even when a strong base or acid is introduced.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation relates the pH of a buffer to the pKa of the weak acid and the ratio of the concentrations of the conjugate base to the acid. It provides a convenient way to calculate the pH and predict how the pH will change upon the addition of acids or bases.

Acid-Base Reaction Stoichiometry

In a buffer, when a strong base is added, it reacts with the weak acid component, converting it into its conjugate base. The stoichiometry of this reaction allows you to determine the changes in the concentrations of the acid and the base, which is crucial for recalculating the pH after the addition.

Buffer Capacity

Buffer capacity refers to the ability of a buffer to neutralize added acid or base without undergoing a significant pH change. It depends on the amounts of the weak acid and conjugate base present in the solution, and it is an important consideration when predicting the effect of adding a strong base or acid.

Transcript

00:01 Some weak acid, hy, partially dissociates into hydrogen ions and y -minus, and this forms a buffer.

00:09 If this buffer has a ph of 8 .77, where h .y has a concentration of 0 .110 molar, and y -minus has a concentration of 0 .220 molar, we can calculate the p .ka of h -y, using a the henderson -hosselbach equation.

00:45 Rearranging this, we get p -k -a equals ph minus the log of the buffer component concentration ratio.

00:58 So now we can plug in our values here.

01:15 This is then equal to 8 .77.

01:19 So minus 0 .301, which is 8 .47.

01:28 So now let's say we add 0 .001.

01:32 105 moles of barium hydroxide to 0 .350 liters of the buffer solution.

01:44 We can find the concentration of barium hydroxide since molarity is just moles over liters.

01:53 So 0 .0015 moles over 0 .350 liters and we get 0 .00429.

02:07 Molar barium hydroxide.

02:15 This means that we have 0 .00429 molar barium and twice as much, so 0 .00857 molar hydroxide.

02:35 This is because there is a one -to -one ratio between barium hydroxide and barium and a one -to -two ratio of barium -hydroxide to hydroxide.

02:45 The weak acid, hy, is going to react with the hydroxide ions from the added barium hydroxide to create water and y minus.

03:07 Hydroxide is a strong base, so this reaction is going to go to completion, which is why there is a single forward -facing arrow.

03:18 So now we can set up an ice table where initially we have 0 .110 ,000, molar, hy, 0 .00857 molar oh -h minus, which is this number here, and 0 .220 molar y -minus.

03:45 Liquids aren't included in these calculations, so we can just ignore water here...

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