Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. I take a weak acid, H A, at ten to the minus three molar, and its K a is also ten to the minus three. Our usual shortcut for the degree of dissociation: alpha is about the square root of K a over c. K a over c is one, and the root of one is one. So this weak acid is one hundred percent ionised? Something has gone wrong, right? Hold that thought; our last example fixes it. This is set three of mixed problems on equilibrium, the kind of numericals that come up often in JEE Main. Two quick ideas, then two full problems. So, let's begin.

0:46Idea 1: how much of an indicator is ionised?

Idea number one. An indicator, written H I n; the I n stands for indicator. Its p K I n is nine: that is capital K with a small I n below it, and the lowercase p in front means minus log. At a p H of eight, what fraction of it is in the I n minus form? The minus up top is its charge. Okay? An indicator is a weak acid, so we use this relation: p H equals p K I n, plus log of I n minus over H I n. The square brackets mean concentration, in moles per litre. Right? Now the numbers. Log of the ratio is p H minus p K I n: eight minus nine, that is minus one. So the ratio is ten to the power minus one, one over ten. With me? So one tenth of the indicator is in the I n minus form, right? No, no, careful! One over ten is the ratio, I n minus to H I n. For every one part of I n minus, there are ten parts of H I n. So the fraction is one over, one plus ten: one over eleven, about nine percent; the wavy equals sign means approximately. Ratio and fraction are different, okay?

2:05Idea 2: two weak acids in one solution

Idea number two. One solution, zero point one molar in H A, and zero point one molar in H B; capital M means molar, moles per litre. K a, capital K with a small a, is the acid dissociation constant: for H A, one times ten to the minus five, and for H B, four times ten to the minus five. Given log five as zero point seven zero, find the p H of the solution. Clear? Now think. Both acids give hydrogen ions, H plus. So whose H plus do we use? Both, exactly! They pour H plus into the same solution, and each acid's K a uses that same total H plus. So, H plus squared equals K a one times c one, plus K a two times c two. The small one and two below label the two acids, and c is each concentration. This works because both acids are weak and ionise only a little, okay? Now the numbers. Ten to the minus five, times zero point one, is ten to the minus six. Four times ten to the minus five, times zero point one, is four times ten to the minus six. Add them: five times ten to the minus six. Right? Take the square root: two point two four times ten to the minus three molar. For the p H, minus log of the root of five times ten to the minus six is half of, six minus log five. Half of six minus zero point seven zero: two point six five. Got it?

3:44Example 1: Ksp of Ca(OH)2 from pH 12.3

Alright, example one. Here, a saturated solution of calcium hydroxide, C a O H two, has a p H of twelve point three. The small two after the bracket means two O H groups. Given log two as zero point three zero, find K s p, the solubility product: capital K with a small s p below it. Pause the video, and try it first. Done? Okay, let's check. It is a base, so go through p O H first. At two hundred and ninety eight kelvin, where K w, the ionic product of water, is ten to the minus fourteen, p H plus p O H is fourteen. Fourteen minus twelve point three: one point seven. Okay? So the hydroxide ion concentration, O H minus, is ten to the power minus one point seven. Write that as ten to the power zero point three, times ten to the minus two. Log two is zero point three, so ten to the zero point three is two. Two times ten to the minus two: zero point zero two molar. Right? Now the calcium ion, C a two plus; the two plus up top is its charge. Is it also zero point zero two molar? No! Each calcium hydroxide gives one calcium ion and two hydroxide ions. So the calcium ion is half of zero point zero two: zero point zero one molar. With me? Now, K s p equals calcium ion, times hydroxide ion squared. Don't forget that square; it comes from the two in the formula. Zero point zero two squared is four times ten to the minus four, and times zero point zero one: four times ten to the minus six. Clear?

5:32Example 2: weak acid with Ka = c, solve the quadratic

Now the harder one: our puzzle from the start. Find the p H of ten to the minus three molar H A, with K a equal to ten to the minus three. Pause the video, and try it. Done? Okay, let's check. The shortcut: alpha, the Greek letter for the degree of dissociation, is about the square root of K a over c. K a over c is one, and the root of one is one: one hundred percent ionised. Can we trust that? No, no, wait! The shortcut assumes the acid ionises only a little, so that c minus x is almost c. The rule: it is fine only when alpha comes out below about five percent. Here K a equals c, alpha comes out as one, so the shortcut fails completely. Okay? So we solve it exactly. Let x be the hydrogen ion concentration; the acid ion, A minus, is also x, and the H A left is ten to the minus three, minus x. So x squared, over ten to the minus three minus x, equals ten to the minus three. Cross multiply, and the double arrow gives a quadratic: x squared, plus ten to the minus three times x, minus ten to the minus six, equals zero. Right? Quadratic formula, with the positive root: minus ten to the minus three, plus the square root of ten to the minus six plus four times ten to the minus six, all over two. That gives x equal to zero point six one eight times ten to the minus three molar. Clear? So the p H is minus log of zero point six one eight times ten to the minus three: three point two one. And the acid is about sixty two percent ionised: a lot, but not one hundred percent. Puzzle solved! Got it?

7:24Quick revision: indicator fraction, two acids, 5% rule

So, quick revision. For an indicator, the ratio r is ten to the power p H minus p K I n, and the fraction in the I n minus form is r over one plus r. Two weak acids together: H plus squared equals K a one c one, plus K a two c two. And for K s p from p H, go through p O H first. Finally, alpha is about root of K a over c only when alpha is below about five percent; otherwise, solve the quadratic. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!