Problem

A 10.0 -g sample of white phosphorus was burned i…

Question

Answered step-by-step

Problem 155 Hard Difficulty

At $28^{\circ} \mathrm{C}$ and 0.982 atm, gaseous compound HA has a density of $1.16 \mathrm{g} / \mathrm{L} .$ A quantity of $2.03 \mathrm{g}$ of this compound is dissolved in water and diluted to exactly 1 L. If the pH of the solution is 5.22 (due to the ionization of $\mathrm{HA}$ ) at $25^{\circ} \mathrm{C},$ calculate the $K_{\mathrm{a}}$ of the acid.

Video Answer

Solved by verified expert

preview

Textbook Answer

Official textbook answer

Video by Susan Hallstrom

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Chemistry 102

Chemistry

Chapter 15

Acids and Bases

Related Topics

Liquids

Discussion

You must be signed in to discuss.

Top Chemistry 102 Educators

Stephanie Castillo

University of Central Florida

Lizabeth Tumminello

Stephen Pulliam

Drexel University

KS

Karli Sipps

Watch More Solved Questions in Chapter 15

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Problem 88

Problem 89

Problem 90

Problem 91

Problem 92

Problem 93

Problem 94

Problem 95

Problem 96

Problem 97

Problem 98

Problem 99

Problem 100

Problem 101

Problem 102

Problem 103

Problem 104

Problem 105

Problem 106

Problem 107

Problem 108

Problem 109

Problem 110

Problem 111

Problem 112

Problem 113

Problem 114

Problem 115

Problem 116

Problem 117

Problem 118

Problem 119

Problem 120

Problem 121

Problem 122

Problem 123

Problem 124

Problem 125

Problem 126

Problem 127

Problem 128

Problem 129

Problem 130

Problem 131

Problem 132

Problem 133

Problem 134

Problem 135

Problem 136

Problem 137

Problem 138

Problem 139

Problem 140

Problem 141

Problem 142

Problem 143

Problem 144

Problem 145

Problem 146

Problem 147

Problem 148

Problem 149

Problem 150

Problem 151

Problem 152

Problem 153

Problem 154

Problem 155

Problem 156

Problem 157

Problem 158

Problem 159

Problem 160

Problem 161

Problem 162

Problem 163

Problem 164

Video Transcript

Let's find K A and in this problem, we're going to be using PV equals NRT. First, we have quite a few givens. Let's get him written down. Our temperature is given as 28 degrees Celsius, which we will need later as 301 k We have a pressure given at 0.982 atmospheres. Um, the asset H A has a density of 1.1 grams per leader. We have a mass of 2.3 grams of h A is going to be dissolved in exactly one leader of solution and a resulting solution has a ph of 5.22 those air givens. So we're to find a k A. This is our unknown. Okay, let's make a plan of how we're going to do this. I'm not too bad here. It's really not too bad of a problem. It takes little time, but to find K A, we need to know polarities. So we're going to use PV equals NRT to find malls. And then once we have moles, we're going to use the density, which is exactly a leader, and end to find Moeller Mass and then we'll use are given mass and molar mass. And the fact that we have exactly one leader that should be a one leader to find polarity that will give us our mill arat e will use after we've got similarity will use ice and we confined values for the ice. You know, things that we're gonna put in for Ph. And once we have all that will be able to substitute into R K equation. Let's get started. Okay. First, to find malls, we're gonna use n equals P B over our teeth and let me get my paper. Here we have 0.9 0.982 atmospheres times exactly one leader I'm saying exactly, cause that won't affect our sig figs. And we're going to use for R 0.8 to 1. L A t m overcame all and 301. Kelvin. When we're done doing our math, we will have for moles. 0.397 moles of H A. Okay, now we're going to figure out our molar mass because we have exactly one leader. So, um, I know that I have 1.16 grams in one leader and I know I have zero point 0397 malls in one later. So I do my math here and I get a molar mass of 29.2 grams per mole. Now, let's use our knowledge that polarity equals moles per leader. And how am I gonna get malls? I know that I had was given 2.3 grams and we'll convert that to malls by using the molar mass. And we have exactly one leader. This will give it 0.695 polarity and that similarity of h A. This is the polarity of H A. Let's do a nice table and we've got a H plus and a So this is zero and zero and this was 0.695 This will increase by negative X plus X plus x leave a little bit more room here and you'll figure out why in a little bit this will be 0.695 minus x X and X. Okay, Now I'm gonna put these values my equilibrium values into my K expression. But I get a lot of excess there, so this isn't going to work too well for me. Do I have a low battery? Let me plugging quick. There we go. So I'm going to use my knowledge that the Ph was 5.22 So if I take 10 to the minus 5.22 I get an H plus concentration of 6.3 times 10 to the minus six Mohler and that value is the value for this X this X and this X. Now we're ready to put R K a equation together. K equals and in my numerator, actually, let me make this a little smaller here, get this on one page. Que is going to be hpe plus and a minus. And those values air the same. They're both acts divided by age A. So we know that we're gonna put an accent. We're gonna put this in for X, So I'm going to have 6.3 times 10 to the minus six squared, divided by 0.695 minus 6.3 times 10 to the minus six. Do the math on this, and your final answer will be 5.2 times 10 to the minus temp. Which means I probably could have ignored that. You can check that if you'd like

Get More Help with this Textbook

Raymond Chang, Kenneth A. Goldsby

Chemistry

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

See More

Related Topics

Liquids

Top Chemistry 102 Educators

Stephanie Castillo

University of Central Florida

Lizabeth Tumminello

Numerade Educator

Stephen Pulliam

Drexel University

KS

Karli Sipps

Numerade Educator