Basicity Calculations and Separations

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take a point one zero molar solution of aniline, a base. Its p H is only about eight point eight. Now take dimethylamine, ten times more dilute. Less base, so a lower p H, right? Not at all, its p H is about eleven point three! How is that possible? Today we will find out. Calculations like these come up often in JEE, so let's begin.

0:30A weak amine base in water

So, we call our amine B, for base. B plus water is in equilibrium with B H plus, the protonated amine, and O H minus, the hydroxide ion. The raised signs are charges, the small two written low is a subscript, and the two half arrows mean equilibrium. Okay? Start with c moles per litre of amine and let x moles per litre react; the square brackets mean concentration. So B H plus and O H minus are both x, and c minus x of the amine is left. The double arrow means implies, so K b, small b written low, is exactly x squared over c minus x. Right? But solving that quadratic is slow in the exam. So let me ask you this: when can we drop the x in c minus x? Yes, exactly, when x is tiny compared to c. The fraction x over c is the degree of ionisation, the Greek letter alpha. If alpha is small, c minus x is approximately c; the wavy equals sign means approximately equal. Got it? Then x, the hydroxide concentration, is the square root of K b times c; the tick shaped sign is the square root. This is the shortcut, but only when alpha is small. With me? Otherwise, when alpha is not small, we solve x squared over c minus x equals K b as a full quadratic. Okay?

1:58The salt of an amine is an acid

Now, the amine with hydrochloric acid gives its salt, B H plus, C l minus, and B H plus is the conjugate acid of the amine. Its constant is K a, where the small letter a stands for acid, and K a equals K w over K b; K w, the ionic product of water, is ten to the minus fourteen at two ninety eight kelvin. Right? Then, for c moles per litre of salt, the H plus concentration is the square root of K a times c, so the solution is acidic. Clear? Now, if the base gets weaker, K b gets smaller. So what happens to K a? It gets larger, yes, exactly, because K b sits below the line. So the weaker the base, the stronger its conjugate acid, and a stronger base pulls that proton away, displacing the weaker base from its salt. Got it?

2:49Amine with its salt: a buffer

Next, an amine mixed with its own salt is a buffer. Its p H equals p K a, plus log of the concentration of B over the concentration of B H plus. The small p means minus log, so p K a is minus log of K a. Free amine on top, salt below, okay? So let me ask you this: if exactly half of the amine is protonated, what is the p H? Half protonated means B and B H plus are equal, so the ratio is one, and log of one is zero. So the p H equals p K a, exactly: the amine is half protonated at p H equal to p K a. Clear?

3:29A stronger base sets a weaker one free

Now let us use this. Here RNH two is an amine, where R is an alkyl or aryl group. With dilute hydrochloric acid, H C l, it forms the salt RNH three plus, C l minus; the arrow means gives. So the amine dissolves in dilute acid. Right? To get it back, we add an alkali such as sodium hydroxide, a much stronger base. It takes the proton, and the free amine comes back, with sodium chloride and water. With me? And that is how amines are separated from neutral and acidic compounds. Shake the mixture, dissolved in ether, with dilute acid: the amine passes into the water layer as its salt, while neutral and acidic compounds stay behind. Got it?

4:15Example 1: pH and degree of ionisation of 0.10 M aniline

Now, Example one, the first half of our puzzle. Find the degree of ionisation and the p H of a point one zero molar solution of aniline, with K b four point two times ten to the minus ten. The capital M means molar, moles per litre. Pause the video and try it first. Done? Okay, let's check. Hydroxide is the square root of K b times c, and K b times c is four point two times ten to the minus ten, times point one zero, which gives four point two times ten to the minus eleven. Write it as forty two times ten to the minus twelve, and the root is about six point five times ten to the minus six molar. Okay? But wait, we used the shortcut. So let me ask you this: is alpha really small here? Alpha is x over c, so six point five times ten to the minus six over point one zero, which gives six point five times ten to the minus five. That is tiny, so the shortcut was fine, and this is the degree of ionisation. Clear? Next, p O H is minus log of the hydroxide concentration, which is five point one nine. At two ninety eight kelvin, p H is fourteen minus p O H, which gives about eight point eight. So the solution is only mildly basic, because aniline is such a weak base. Right?

5:38Example 2: pH of 0.010 M dimethylamine by the quadratic

Now, Example two, the second half of our puzzle. Find the p H of a point zero one zero molar solution of dimethylamine, with K b five point four times ten to the minus four. Pause the video and try it first. Done? Okay, let's check. Most students take the square root of K b times c straight away. So let me ask you this: is that safe here? Careful now, because this is the trap. K b times c is five point four times ten to the minus six, and its square root is about two point three times ten to the minus three. That is twenty three percent of c, not small at all, so no shortcut here. Okay? So we solve the exact equation, x squared over point zero one zero minus x, equals K b. Multiplying across gives x squared, plus five point four times ten to the minus four, times x, minus five point four times ten to the minus six, equals zero. Right? Use the quadratic formula and keep the positive root, as a concentration is never negative. So x, the hydroxide concentration, is two point zero seven times ten to the minus three molar, which is about twenty one percent of c, okay? Then p O H is minus log of that, which is two point six eight. So the p H is fourteen minus two point six eight, which is about eleven point three. Got it? Now look at the reason. The K b of dimethylamine is over a million times that of aniline, four point two times ten to the minus ten. So even ten times more dilute, it gives the higher p H, which solves our puzzle. Clear?

7:18Quick revision: weak base, salt, buffer, displacement

So, quick revision. For a weak base, hydroxide is the square root of K b times c, only when alpha is small; otherwise, solve the quadratic. For a salt, K a is K w over K b, and H plus is the square root of K a times c; for a buffer, p H is p K a plus log of B over B H plus. And a stronger base frees a weaker one from its salt. Now try the practice problems on this page. All the best!