Weak Acids and Bases: Kₐ, K_b and Ostwald's Dilution Law

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. I take a weak acid, like the acetic acid in vinegar, and dilute it one hundred times. Now a much bigger fraction of it ionises. So, more ionisation, a more acidic solution, right? Hmm, think again. By the end of this video, you will know the answer. On the way, we learn K a, K b, the p H of weak acids and bases, and Ostwald's dilution law. These come up often in JEE, so let's begin.

0:34Weak acids: only a small fraction ionises

So, take a weak acid, H A, okay? It gives H plus and A minus; the small plus and minus signs up top are charges. The two half arrows, pointing both ways, mean equilibrium. Its concentration is c, and alpha, this Greek letter, is the fraction that ionises: the degree of ionisation. Right? Now tell me. At equilibrium, how much H plus, how much A minus, and how much H A is left? Yes, exactly! c times alpha ionises, so H plus and A minus are each c alpha; the square brackets mean concentration. And the H A left over is c times, one minus alpha. Clear? Now, K a, a capital K with a small a below, is the acid ionisation constant: H plus times A minus, over H A, okay? Put in our values, cancel one c, and you get c alpha squared, over one minus alpha. The small two means squared. Got it? And a weak base works the same way. Base B takes an H plus from water, H two O, the small two meaning two hydrogen atoms, and gives B H plus and O H minus. Its K b, small b for base, has exactly the same form. With me?

1:49When α is small: the working formulas

Now, for a weak acid, alpha is often tiny. So, one minus alpha is close to what? One, correct! But only when alpha is less than, or about equal to, five percent; that is the less than sign with a wavy line below. And the wavy equals sign means approximately equal. Okay? So K a becomes just c alpha squared. The double arrow means, so this gives: alpha is about the square root of K a over c. Right? Then H plus is c times alpha, the square root of K a times c. Take minus log, and here is a topper's shortcut: p H equals half of, p K a minus log c. p K a is minus log of K a, with log to the base ten. Clear? For a weak base, the same story: O H minus is root K b times c, and p O H is half of, p K b minus log c. Okay?

2:47Ostwald's dilution law

Now look at alpha again: root K a over c. This is Ostwald's dilution law. Dilute, and c goes down, the down arrow; so alpha goes up, the up arrow. Right? Now, our puzzle. Dilute one hundred times. What happens to alpha? Ten times larger, yes! c becomes one hundredth, and the root of a hundred is ten. So alpha becomes ten times bigger; the cross means times. Got it? So, ten times more ionised. More H plus, more acidic, right? No, no, careful! H plus is root K a times c, so when c falls a hundred times, H plus falls ten times; the line with two dots is the division sign. Ten times less H plus means the p H rises by one. Clear? So, puzzle solved: on dilution, the acid is more ionised, yet less acidic. H plus always falls on dilution; the exact factors, ten and one, hold while alpha stays small, where our formula works. With me?

3:50Conjugate pairs and the common ion

Now, one more link. For an acid and its conjugate base, which differ by one H plus, K a times K b equals K w, the ionic product of water. Okay? Take minus log: p K a plus p K b equals fourteen, at two hundred ninety eight kelvin, that is twenty five degrees Celsius, where K w is ten to the minus fourteen. Quick: p K a of acetic acid is four point seven four. What is p K b of acetate? Nine point two six, correct! Fourteen minus four point seven four. Right? Last, the common ion. If a strong acid like H C l is also present, at concentration capital C, its H plus pushes the weak acid's equilibrium back, by Le Chatelier's principle. So alpha is about K a over capital C, much smaller than in water alone. Got it?

4:40Example 1: pH and alpha of 0.10 M acetic acid

Now, Example one. Find the H plus concentration, the p H, and the degree of ionisation of zero point one zero molar acetic acid; capital M means molar, moles per litre. K a is one point eight times ten to the minus five. Pause the video, and try it first. Done? Okay, let's check. Acetic acid is weak, so H plus is not zero point one. Use root K a times c: root of, one point eight times ten to the minus five, times zero point one zero. Okay? That is root of one point eight times ten to the minus six, which gives one point three four times ten to the minus three molar. Right? So, p H is minus log of that: two point eight seven. With me? Check with the topper's shortcut. p K a is four point seven four, and minus log of zero point one is plus one. Half of five point seven four: two point eight seven again! See how fast? Finally, alpha is H plus over c: one point three four percent. Below five percent, so the shortcut was safe.

5:53Example 2: pH of 0.010 M HF, when the shortcut fails

Clear? Now a harder one. Find the p H of zero point zero one zero molar H F, hydrofluoric acid, with K a six point eight times ten to the minus four. Pause and try. Hint: check alpha first. Done? Same as before, root K a times c, and finish, right? No, no, wait! Check alpha first. Root of K a over c is root of zero point zero six eight: twenty six percent. Far above five percent, so the shortcut fails. Okay? So, solve exactly. Let x be the H plus concentration; F minus is also x, and the H F left is zero point zero one zero minus x. So, x squared over zero point zero one zero minus x equals K a. Right? Multiply across: x squared, plus six point eight times ten to the minus four times x, minus six point eight times ten to the minus six, equals zero. It is a quadratic. Got it? Keep the positive root: x is two point two nine times ten to the minus three molar. So, p H equals two point six four. With me? And alpha is twenty three percent, not twenty six. So, whenever alpha comes out above about five percent, solve the quadratic. That's it!

7:13Quick revision: Ka, Ostwald's law and pKa + pKb = 14

So, quick revision. For a weak acid, K a is c alpha squared over one minus alpha; and when alpha is below about five percent, alpha is root K a over c, and H plus is root K a times c. On dilution, alpha rises, but H plus falls, and the p H rises. p K a plus p K b is fourteen at two hundred ninety eight kelvin, and if alpha is above five percent, solve the quadratic. Now go and try the practice problems on this page. All the best!