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