(a) Calculate the pH of a buffer that is 0.125 M in NaHCO3 and 0.095 M in Na2 CO3 . (b) Calculat

(a) Calculate the pH of a buffer that is 0.125 M in NaHCO3...

Key Concepts

Stoichiometric Calculations in Mixing Solutions

When preparing buffer solutions by mixing different volumes and concentrations of compounds, stoichiometric calculations are essential to determine the final concentrations of the buffer components. These calculations ensure the accurate prediction of the system’s pH after mixing.

Conjugate Acid-Base Pairs

Conjugate acid-base pairs are related species that differ by one proton. In a buffer, the weak acid and its conjugate base work together to maintain pH stability by shifting the equilibrium in response to added acids or bases.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation provides a straightforward way to calculate the pH of a buffer solution using the pKa of the weak acid and the ratio of the concentrations of the conjugate base to the weak acid. This equation is an indispensable tool in understanding and designing buffer systems.

Buffer Solution

A buffer solution is a mixture that contains a weak acid and its conjugate base (or a weak base and its conjugate acid) which work together to stabilize the pH of the solution when small amounts of acid or base are added. Buffers resist changes in pH, making them essential in both chemical reactions and biological systems.

Transcript

00:01 In this problem, we're asked to calculate the ph of a buffer solution consisting of sodium bicarbonate and sodium carbonate.

00:09 We have two calculations to do.

00:11 Our first calculation, we are asked to find the ph of a sub buffer that is 0 .150 molar.

00:21 I lied.

00:22 Wrong problem.

00:26 0 .125 molar in sodium bicarbonate and 0 .1 .5 molar and sodium bicarbonate and 0 .1 .5 molar and 0 .5 .5 molar in sodium bicarbonate.

00:37 0 .095 molar in sodium carbonate.

00:47 In order to do this, like everything, i like to write my chemical equation, which i will in a moment.

00:58 What else? the conjugate acid for this buffer, here's a note, will be my h -c -o -3, my bicarbonate ion.

01:20 And the bicarbonate ion has a k -a of 5 .6 times 10 to...

01:26 The minus 11th.

01:33 I'm going to write that again here.

01:34 5 .6 times 10 to the minus 11th.

01:38 And that will be my ka for this one.

01:40 It's actually the ka2 for carbonic acid.

01:46 Okay, what else shall we do here? let's go ahead and i'll just switch colors for fun.

01:54 If we want to write our chemical equation for this one, it's going to be h.

02:03 C .o .3.

02:12 We'll have an h.

02:13 Plus.

02:15 And co3 to minus.

02:22 And my initial concentrations will be as follows.

02:29 This one is 0 .125 molar, 0 and 0 .095.

02:40 That's my initial.

02:42 My change will be minus x plus x and plus x.

02:50 And my equilibrium concentrations will be 0 .125.

02:54 Minus x, which, of course, i'm sure you figured out, i will be ignoring my plus x and minus x because it will be insignificant or will have negligible impact on my result.

03:11 And what shall we do next? let's write our ka expression, equal our h plus concentration, times our nion concentration, primary anion, and my conjugate acid concentration.

03:29 Concentration and substituting.

04:01 So let me go ahead and highlight those two portions that we're going to omit.

04:10 And in solving for h plus here, i'm going to change my x to h plus.

04:17 That will equal 5 .6 times 10 to the minus 11th times 0 .125 divided by 0 .095.

04:34 So my that will equal switch to a brighter.

04:39 That will equal.

04:40 An h plus concentration of where is my answer? 7 .36 times 10 to the minus 11th.

04:56 And of course the negative log of 7 .36 times 10 to the minus 11th is 10 .13 as my ph.

05:10 That is the answer for part 8.

05:13 Part b, similar calculation...

Please give Ace some feedback