There is a buffer system (H2 PO4^--HPO4 ^2-) in blood that helps keep the blood pH at about 7.40

step 1 So we have a buffer, which is H2. PO4 minus as the weak acid, and its conjugate base is HPO2 -2

There is a buffer system (H2 PO4^--HPO4 ^2-) in blood that...

There is a buffer system $\left(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}-\mathrm{HPO}_{4}{ }^{2-}\right)$ in blood that helps keep the blood $\mathrm{pH}$ at about $7.40 .\left(\mathrm{K}_{\mathrm{a}} \mathrm{H}_{2} \mathrm{PO}_{4}^{-}=6.2 \times 10^{-8}\right)$ (a) Calculate the $\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right] /\left[\mathrm{HPO}_{4}^{2-}\right]$ ratio at the normal $\mathrm{pH}$ of blood. (b) What percentage of the $\mathrm{HPO}_{4}{ }^{2-}$ ions are converted to $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$ when the $\mathrm{pH}$ goes down to $6.80 ?$ (c) What percentage of $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$ ions are converted to $\mathrm{HPO}_{4}{ }^{2-}$ when the $\mathrm{pH}$ goes up to $7.80 ?$

There is a buffer system $\left(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}-\mathrm{HPO}_{4}{ }^{2-}\right)$ in blood that helps keep the blood $\mathrm{pH}$ at about $7.40 .\left(\mathrm{K}_{\mathrm{a}} \mathrm{H}_{2} \mathrm{PO}_{4}^{-}=6.2 \times 10^{-8}\right)$
(a) Calculate the $\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right] /\left[\mathrm{HPO}_{4}^{2-}\right]$ ratio at the normal $\mathrm{pH}$ of blood.
(b) What percentage of the $\mathrm{HPO}_{4}{ }^{2-}$ ions are converted to $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$ when the $\mathrm{pH}$ goes down to $6.80 ?$
(c) What percentage of $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$ ions are converted to $\mathrm{HPO}_{4}{ }^{2-}$ when the $\mathrm{pH}$ goes up to $7.80 ?$

Key Concepts

Percent Conversion in Equilibrium Systems

Percent conversion involves calculating the fraction or percentage of a substance that is converted from one form to another, based on the equilibrium established in an acid-base system. This calculation leverages the equilibrium constant and the pH value to determine the extent of the reaction, reflecting how changes in pH affect the balance between ionized and unionized forms.

pKa and Acid Dissociation Constant

The pKa is the negative logarithm of the acid dissociation constant (Ka) and serves as an indicator of the strength of a weak acid. Understanding pKa is crucial because it informs how completely an acid will dissociate at a given pH, which in turn determines the ratio of the acid to its conjugate base in a buffer solution.

Henderson-Hasselbalch Equation

This equation provides a direct relationship between the pH of a buffer solution, the acid dissociation constant (pKa), and the ratio of the concentrations of conjugate base and acid. It is a fundamental tool in calculating the proportions necessary for a buffer to maintain a desired pH, and it simplifies the analysis of acid-base systems by relating the pH to the logarithm of the concentration ratio.

Acid-Base Equilibrium

Acid-base equilibrium describes the balance between an acid and its conjugate base in solution, establishing a reversible reaction where the acid donates a proton and the base accepts it. This equilibrium is characterized by the acid dissociation constant (Ka), and its understanding is critical for predicting the direction of shifts in pH under varying conditions.

Buffer Solutions

Buffer solutions consist of a weak acid and its conjugate base, or a weak base and its conjugate acid, working together to stabilize the pH of a solution. They resist changes in pH upon the addition of small amounts of acid or base by absorbing the excess H+ or OH– ions, making them essential in many biological and chemical systems.

Transcript

00:01 So we have a buffer, which is h2.

00:04 Po4 minus as the weak acid, and its conjugate base is hpo2 -2 minus.

00:11 So the k .a for our weak acid is 6 .2 times 10 to negative 8, and we know the ph is going to be 7 .4.

00:23 So that means our h -plus concentration, 10 to the negative 7 .4, or 3 .98 times 10 of the negative 8 molar.

00:33 So first thing we're going to do is find the ratio of acid to base.

00:39 So for a buffer, h plus concentration equals ka times the concentration of your acid over the concentration of your base.

00:52 So our h plus we said was 3 .98 times 10 to the negative 8.

01:00 And we'll plug in ka here.

01:04 So that'll give us a ratio of our acid to base.

01:19 So i'll give us a ratio of 0 .64.

01:26 And then what we're going to do is we're going to lower the ph to 6 .7 by converting some of the base to acid.

01:37 Okay, so ph is now 6 .80.

01:45 That means our new h plus concentration is 1 .6 times down the negative 7 molar.

01:53 So 1 .6 times 10 and negative 7 equals ka, which isn't not going to change.

02:06 Now, we're starting with 0 .64 moles of acid for every one mole of base, and we're told we're going to convert some of this base, so minus x and plus x.

02:19 Because when the base is converted, it's converted to weak acid.

02:25 So however much weak base we lose, we gain that much.

02:27 Weak acid.

02:29 So we'll just go ahead and figure out the math here.

02:33 2 .58 is 0 .64 plus x over 1 minus x...

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