Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Take hydrochloric acid, a strong acid, and dilute it until it is ten to the power minus eight molar. p H is minus log of the H plus concentration, so the p H is eight, right? But p H eight is basic. An acid, giving a basic solution? Hmm, something is wrong. Example two will settle it. Today is a concept check class: six true or false statements from the whole equilibrium unit. Such statements come up often in JEE questions. So, let's begin.

0:37Check 1: does a catalyst change Kc?

So, check one. Adding a catalyst changes K c: capital K with a small c below, the equilibrium constant in concentrations. Now, a catalyst makes a reaction faster, so surely it gives more product, right? True or false? Let's see what a catalyst really does. It lowers the activation energy, the energy barrier, of the forward and the backward reactions, by the same amount. Both directions, okay? So both rates rise by the same factor. Equilibrium is reached sooner, and that is all. With me? So the equilibrium composition does not change, and K c does not change. The statement is false! Clear? Now write this line down: capital K depends only on temperature, not on concentration, pressure, volume or a catalyst. We will need it again today. Right?

1:31Check 2: ammonium chloride in ammonia

Next, check two. Adding N H four C l, ammonium chloride, to aqueous ammonia lowers the p H; the small four below is a subscript. Does the p H really go down? True or false? Start from the equilibrium. Ammonia, N H three, plus water, H two O, gives ammonium, N H four plus, and hydroxide, O H minus. The plus and minus up top are charges; the two half arrows mean equilibrium. Okay? Now ammonium chloride gives extra N H four plus, the very ion already on the right side. We call it the common ion. So, which way does the equilibrium shift? Backward, yes, exactly! Le Chatelier's principle. The common ion suppresses the ionisation of ammonia, so the hydroxide concentration falls; square brackets mean concentration. Right? Fewer hydroxide ions, so the solution is less basic, and the p H falls. So this one is true! Careful, not every statement today is false. Got it?

2:33Check 3: diluting a buffer

Now, check three. Diluting a buffer tenfold changes its p H by one unit. That idea comes from strong acids: zero point zero zero one molar H C l, capital M for molar, has p H three; diluted tenfold, ten to the power minus four molar, p H four. So a buffer also moves one unit. True or false? No, no, careful! The buffer is not a strong acid. Its p H comes from the Henderson equation: p H equals p K a, plus log of salt over acid. p K a is minus log of K a, the acid ionisation constant. Okay? Now dilute it. Salt and acid are both diluted ten times. So what happens to their ratio? Nothing at all! Very good! Both fall ten times, so the ratio is unchanged, and so is the log term. Clear? So the p H stays the same, or nearly the same, as the brackets say. The statement is false. But the buffer capacity falls: it can now absorb less acid or base. With me?

3:39Check 4: Ksp and a common ion

Now check four. K s p of A g C l, silver chloride, is smaller in zero point one molar N a C l, sodium chloride, than in water. K s p, capital K with a small s p below, is the solubility product. True or false? And think: what exactly gets smaller? Solid silver chloride, the small s in brackets, gives silver ions and chloride ions; a q in brackets means aqueous. K s p equals the silver ion concentration times the chloride ion concentration. Okay? In sodium chloride, the chloride concentration is already large: chloride is the common ion. The product must still equal K s p, so the silver ion concentration must be small. Right? So less silver chloride dissolves. The solubility, small s, becomes about K s p over c, where c is the chloride concentration; the wavy equals sign means approximately. Smaller solubility, yes! But look at K s p itself: it does not change. It depends only on temperature. At a fixed temperature, it is the same in water and in sodium chloride, so the statement is false. Mixing up solubility and K s p is a classic trap. Got it?

4:55Example 1: is Kp = Kc for PCl5 dissociation?

Okay, example one. Phosphorus pentachloride, P C l five, splits into phosphorus trichloride, P C l three, and chlorine, C l two. True or false: K p equals K c? K p is the equilibrium constant in partial pressures, the small p below. Pause the video, and try it first. Done? Okay, let's check. For ideal gases, K p equals K c, times R T to the power delta n g; R is the gas constant, capital T the temperature in kelvin. The little triangle is the capital Greek letter delta, meaning change, and the small g means gas: gaseous moles only. So, what is delta n g here? Products: one mole of P C l three plus one mole of chlorine, two moles of gas. Reactants: one mole of P C l five. So delta n g is one plus one, minus one: one. Right? So K p equals K c times R T, and that is not K c; the slashed equals sign means not equal. The statement is false! Clear? K p equals K c only when delta n g is zero, as in hydrogen plus iodine giving two H I: two moles of gas on each side. Always count the gas moles. Okay?

6:13Example 2: pH of 10^-8 M HCl, true or false?

Now example two, the harder one, and it is our opening puzzle. The p H of ten to the power minus eight molar H C l is eight. True or false? Pause and try. Hint: water ionises too. Done? Okay, let's check. Minus log of ten to the minus eight is eight, but p H eight is basic! At twenty five degrees Celsius, the small circle meaning degrees, an acid can never give a p H above seven, however dilute. So the statement is false. But what did the simple method forget? Water itself, yes! At such a tiny acid concentration, water's own H plus matters. Let the total H plus be x. The acid gives ten to the minus eight; water gives the rest, and an equal amount of O H minus. So O H minus is x minus ten to the minus eight. Okay? Now K w, the ionic product of water, is ten to the power minus fourteen at twenty five degrees Celsius. So x, times x minus ten to the minus eight, equals ten to the minus fourteen. Right? Open the bracket: x squared, minus ten to the minus eight times x, minus ten to the minus fourteen, equals zero. The double arrow means "so". Take the positive root: x equals one point zero five times ten to the minus seven molar. With me? So p H is minus log of that: six point nine eight. Just below seven: slightly acidic. Puzzle solved! So, when an acid is very dilute, always include the H plus from water. Got it?

7:50Quick revision: what changes K, buffers, dilute acids

So, quick revision. One: K c, K p and K s p change only with temperature, never with a catalyst or a common ion. Two: a common ion suppresses ionisation; a buffer's p H survives dilution, its capacity does not. Three: K p equals K c only when delta n g is zero; and for a very dilute acid, include water's H plus. Now go and try the practice problems on this page. All the best!