1. what is the concentration of acetic acid, given it takes 22 ml of .2 m naoh to neutralize a 52 ml sample of the acids?

2. if 75 ml of 2.8 m koh is needed to neutralize a 35 ml sample of nitric acid, what is the concentration of the acid?

3. what volume of 5 m naoh is needed to neutralize 200 ml of 3.0 m hcl?

4. given that 40 ml of sulfuric acid are neutralized 22 ml of .6m ca(oh)2, what is the concentration of the acid?

5. what is the ph of a solution that has a [h+] concentration of 1 x 10-5?

6. what is the ph of a solution that has a [oh-] concentration of 1 x 10-8?

7. what is the ph of a solution that has a [h+] concentration of 2.5 x 10-4?

8. what is the [h+] concentration of a solution with a ph of 12?

9. what is the [oh-] concentration of a solution with a ph of 10?

1) [CH3COOH]= 0.085 M

2) [HNO3]= 6 M

3) volume of HCl = 0.12 L = 120 mL

4) [H2SO4] = 0.33M

5) pH = 5

6) pH = 6

7) pH = 3.6

8) [H+] =  10^-12 M

9) [OH-] = 10^-4 = 0.0001 M

Explanation:

1. What is the concentration of acetic acid, given it takes 22 mL of 0.2 M NaOH to neutralize a 52 mL sample of the acids?

CH3COOH + NaOH → CH3COONa + H2O

C1V1 = C2V2

⇒ with C1 = Concentration of CH3COOH

⇒ with V1 = 52 mL = 0.052 L CH3COOH

⇒ with C2 = concentration of NaOH = 0.2 M

⇒ with V2 = the volume of NaOH = 22 mL = 0.022 L

C1 = (C2V2)/V1

C1 = (0.2 * 0.022)/0.052

C1  = [CH3COOH]= 0.085 M

2. If 75 mL of 2.8 M KOH is needed to neutralize a 35 mL sample of nitric acid, what is the concentration of the acid?

KOH + HNO3 → KNO3 + H2O

C1V1 = C2V2

⇒ with C1 = Concentration of KOH = 2.8 M

⇒ with V1 = 75 mL = 0.075 L KOH

⇒ with C2 = concentration of HNO3 = ?

⇒ with V2 = the volume of HNO3= 35 mL = 0.035 L

C2 = (C1V1)/V2

C2 = (2.8 * 0.075)/0.035

C2 = [HNO3]= 6 M

3. What volume of 5 M NaOH is needed to neutralize 200 mL of 3.0 M HCl?

NaOH + HCl → NaCl + H2O

C1V1 = C2V2

⇒ with C1 = Concentration of NaOH = 5.0 M

⇒ with V1 = volume of NaOH = ?

⇒ with C2 = concentration of HCl = 3.0 M

⇒ with V2 = the volume of HCl= 200 mL = 0.200 L

V1 = (C2*V2)/C1

V2 = (3.0 * 0.200)/5.0

V2 = volume of HCl = 0.12 L = 120 mL

4. Given that 40 mL of sulfuric acid are neutralized 22 mL of .6M Ca(OH)2, what is the concentration of the acid?

H2SO4 + Ca(OH)2 → CaSO4 + 2H2O

C1V1 = C2V2

⇒ with C1 = Concentration of H2SO4 = ?

⇒ with V1 = volume of H2SO4 = 40.0 mL = 0.04 L

⇒ with C2 = concentration of Ca(OH)2 = 0.6 M

⇒ with V2 = the volume of Ca(OH)2 = 22 mL = 0.022 L

C1 = (C2V2)/V1

C1 = (0.6*0.022)/0.04

C1 = [H2SO4] = 0.33M

5. What is the pH of a solution that has a [H+] concentration of 1 x 10-5?

pH = -log [H+]

pH = -log (1*10^-5)

pH = 5

6. What is the pH of a solution that has a [OH-] concentration of 1 x 10-8?

pH = 14 - pOH

pOH = -log [OH-]

pOH = 8

pH = 14-8 = 6

7. What is the pH of a solution that has [H+] concentration of 2.5 x 10-4?

pH = -log [H+]

pH = -log (2.5 * 10^-4)

pH = 3.6

8. What is the [H+] concentration of a solution with a pH of 12?

pH = -log[H+]

12 = -log[H+]

[H+] = 10^-12 M

9. What is the [OH-] concentration of a solution with a pH of 10?

pOH = 14 - pH

pOH = 14 -10 = 4

pOH = -log[OH-]

4 = -log[OH-]

[OH-] = 10^-4 = 0.0001 M

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.