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6 August, 07:10

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?

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Answers (2)
  1. 6 August, 07:15
    0
    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
  2. 6 August, 07:27
    0
    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
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