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) 0.84 M

2) 6 M

3) 120 ml

4) 3.3 M

5) pH = 5

6) pH = 6

7) pH = 3.6

8) [H+] = 1x10^-10

9) [OH+] = 1x10^-4

Explanation:

1) M x V = M' x V'

so 2 x 22 = M' x 52

M' = 0.84 M

2) M x V = M' x V'

so 2.8 x 75 = M' x 35

M' = 6 M

3) M x V = M' x V'

so 5 x V = 3 x 200

V = 120 ml

4) M x V = M' x V'

so M x 40 = 6 x 22

M = 3.3 M

5) pH = - log [H+]

= - log [ 1x10^-5] = 5 (you can solve it direct if you just look to 5 in top of x10)

6) pOH = - log [OH-]

= - log [ 1x10^-8] = 8

* pH + pOH = 14

so pH = 14 - pOH

= 14 - 8 = 6

7) pH = - log [H+]

= - log [ 2.5x10^-4 ]

= 3.6

8) pH = - log [H+]

12 = - log [H+]

[ H+] = 1x10^-12

9) pH = - log [H+]

10 = - log [H+]

[H+] = 1x10^-10

** * [H+] x [ OH - ] = 10^-14

so [ 1x10^-10 ] x [ OH - ] = 10^-14

[ OH-] = 10^-14 / 10^-10 = 1x10^-4

* * you can after you have [H+] = 1x10^-10

that is mean pH = 10 so pOH will be = 4

so [OH-] = 1x10^-4

Good Luck

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