# CHEM126 – CHAPTER 16 – BASIC HOMEWORK / ACIDS I

## ACIDS I

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### Question 1

What is the pOH of a solution where the [OH] =7.74 x 10-3 M?
REPORT IT TO 2 DECIMAL PLACES.

pOH = -log[OH]

### Explanation

pOH = -log(7.74e-3) = 2.11

### Question 2

What is the pOH of a solution where the [H+] =3.71 x 10-11 M? REPORT IT TO 2 DECIMAL PLACES.

### Explanation

pH= -log(3.71e-11) = 10.43

pOH + pH = 14

pOH= 14 – 10.43

=3.57

### Question 3

The pH is 1.37. What is the pOH?

### Explanation

pOH + pH = 14

pOH = 14 – 1.37 = 12.63

### Question 4

The pOH is 0.94. What is the pH?

### Explanation

pOH + pH=14

pH= 14 – 0.94 = 13.06

### Question 5

A solution with 6.5 mol HCl is mixed with a solution with 11.5 mol of NaOH. How many moles of OH remain?

Acid-base neutralization

### Explanation

11.5-6.4 = 5 mol OH-

### Question 6

A solution with 25.3 mol HCl is mixed with a solution with 11.8 mol of NaOH. How many moles of H+ remain?

### Explanation

Excess= HCl = 25.3-11.8 = 13.5 mol H+

### Question 7

A solution containing 42.3 mol HCl is mixed with a solution containing 35.0 mol NaOH. The volume is 348 L. What is the final pH?

### Explanation

Excess= HCl = 42.3 – 35 = 7.3 mol / 348 L = 0.020977 [H+]

pH= -log [0.020977] = 1.68

### Question 8

A solution with 44.8 mol HCl is mixed with a solution with 47.8 mol NaOH. The volume is 844 L. What is the final pH?

Neutralization and pH example 2

### Explanation

Excess = NaOH  = 47.8 – 44.8 = 3 mol/ 844 L = 0.00355 [OH-]

pOH= -log[0.00355]=2.45

14-2.45=11.55

### Question 9

How many moles of H+ are in 203 mL of 0.52 M nitric acid, HNO3?

Calculating moles

### Explanation

0.203 L (0.52 M) = 0.10556 mol [H+]

### Question 10

How many moles of OH are in 758 mL of 0.42 M NaOH?

### Explanation

0.758 L (0.42 M) = 0.31836 mol [OH-]

### Question 11

How many moles of OH are in 528 mL of 0.26 M Ca(OH)2?

### Explanation

0.528 L ( 0.26 M) = 0.27456 [OH-]

### Question 12

When 132 mL of acid is mixed with 480 mL of base, what is the new volume in L?

### Explanation

0.132 + 0.480 = 0.612 L

### Question 13

439 mL of 0.500 M HCl is mixed with 217 mL of 0.380 M NaOH. What is the pH?

### Explanation

(0.439 L) (0.5 M) = 0.2195 mol H+

(0.217 L) ( 0.38 M)= 0.08246 mol OH-

[H+] – [OH-]= 0.13704 mol / 0.656 L = 0.2089 [H+]

pH= -log[0.2809]= 0.680

### Question 14

205 mL of 0.330 M HBr is mixed with 240 mL of 0.570 M KOH. What is the pH?

### Explanation

(0.205 L) (0.33 M) = 0.06765 mol H+

(0.240 L) (0.57 M) = 0.1368 mol OH-

[OH-] – [H+] = 0.06915 mol OH- / 0.445 L = 0.155 M [OH-]

pOH = -log[OH-] = 0.80857

14-pOH=pH

pH= 13.19

### Question 15

2.0 mL of 0.110 M HBr is mixed with 61.5 mL of 0.62 M HCl. What is the molarity of the H+?

### Explanation

(0.062 L)(0.110 M) = 0.00682 mol H+

(0.0615 L)(0.62 M) = 0.03813 mol H+

[H+]= (0.00682 + 0.03813)/ (0.62 + 0.0615) = 0.364

### Question 16

21.7 mL of 0.630 M NaOH is mixed with 53.4 mL of 1.90 M LiOH. What is the molarity of the OH?

### Explanation

(0.0217 L)(0.63 M) = 0.013671 mol OH-

(0.0534 L)  (1.9 M) = 0.10146 mol OH-

0.115131 mol OH- / (0.0751 L) = 1.533 [OH-]

### Question 17

A solution contains 0.0247 M HBr and 0.150 M acetic acid. What is the pH?

### Explanation

(acetic acid is negligible (doesn’t count) because  it’s hella weak in comparison to [H+])

pH= -log[0.0247] = 1.607

### Question 18

A solution contains 0.0172 M HNO3, 0.0341 M HI, and 0.220 M formic acid, HCOOH. What is the pH?

### Explanation

(formic acid is also hella weak)

pH= -log[0.0172 + 0.0341] = 1.2899

### Question 19

A solution contains 0.137 M Ba(OH)2 (strong base) and 0.140 M ammonia, NH3 (weak base). What is the pH?

### Explanation

pOH = -log(2 * 0.137) = 0.562

pH= 14-0.5622= 13.44