What is the pH of a liter citrate buffer solution containing 0.3 M citric acid and 0.2 M potassi

What is the pH of a liter citrate buffer solution containing...

Key Concepts

Conjugate Acid-Base Pair

A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton. In the context of buffer solutions, the equilibrium between the weak acid and its conjugate base is critical for maintaining pH. The ability of the conjugate base to accept protons when acids are added, and for the weak acid to donate protons when bases are added, is what gives buffer solutions their characteristic pH stability.

Acid Dissociation Constant (Ka) and pKa

The acid dissociation constant, Ka, measures the extent to which a weak acid dissociates into its conjugate base and a proton in solution. The pKa is the negative logarithm of the Ka and provides a more convenient scale to express the acid strength. A lower pKa value indicates a stronger acid because it implies a higher degree of dissociation. Understanding Ka and pKa is essential for adjusting and predicting the behavior of buffer solutions.

Buffer Solution

A buffer solution is a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid) that resists changes in pH when small amounts of acid or base are added. Buffers work by neutralizing added acids or bases, thereby maintaining a relatively stable pH environment, which is crucial in many chemical and biological processes.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation provides a simple way to estimate the pH of a buffer solution. It relates the pH to the pKa of the weak acid and the ratio of the concentrations of the conjugate base to the weak acid. The equation is given by pH = pKa + log([A?]/[HA]), and its utility lies in its ability to predict how variations in the acid or base concentrations affect the pH of the solution.

Transcript

00:03 Problem 67.

00:07 Accrodding to table 16 .3 citric acid, acquiesceded into hydrogen ion aquas plus citrate ion minus aquas.

00:54 Now, at initially, concentration of citric acid is now at equilibrium concentration of citric acid is let x mole of hydrogen ion formed at equilibrium so x moles of citrate ion is also formed then the remaining mole of citric acid at equilibrium is 0 .040 minus x m now we can write k a 1 for this reaction equals to concentration of product that is concentration of hydrogen ion multiply by concentration of citrate ion divided by concentration of citric acid that is concentration of oh okay now put the values so x is square divided by 0 .040 minus x and value of k a 1 for according to table 16 .3 is 7 .4 into 10 to the power minus 4 now we can assume this x is a very small as compared to initial concentration because syphic acid is a weak acid so less dissociated now we can write x square equals to 0 .0 40 minus x sorry 0 multiply by 7 .4 into 10 to the power minus 4.

03:15 After calculating we get value of x equals to 5 .4 into 10 to the power minus 3m.

03:25 Since the value of x is rather large in relation to 0 .050, for a better approximation for x can be obtained by substituting the first estimate into expression for x2.

03:41 Then solving again for x.

03:45 Now we can solve again x.

03:48 So we can write x square equals to 0 .040 minus x into 7 .4 into 10 to the power minus 4.

04:08 That is equal to now we can substitute the value of x is here.

04:17 So, 0 .040 minus value of axis, this 5 .4 into 10 to the power minus 3 into 7 .4 into 10 to the power minus 4.

04:36 Now, x squared equals to, after calculating, we get 2 .557 into 10 to 10 to the power minus 4.

04:49 5 and value of x equals to 5 .1 into 10 to the power minus 3m now this is the result same result obtained from the quadratic formula okay this is a hydrogen concentration from value of k now the the correction the correction to value of x throw not large is significant does the second ionization product is a significant additional concentration of hydrogen ion so we need to check therefore let we can write here okay the correction to the value of x through not large is significant okay so does the second ionization product is significant fecant additional concentration of hydrogen ion? so we need to check.

06:53 So we again write for second part...

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