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How can 100ml of sodium hydroxide solution with a ph of 13.00 be converted to a sodium hydroxide solution with a ph of 12.00?

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Dejanras
Chemical reaction: NaOH(aq) → Na⁺(aq) + OH⁻(aq).
V₁(NaOH) = 100 mL ÷ 1000 mL/L = 0.1 L.
pOH = -log[OH⁻].
pH + pOH = 14.
pOH₁ = 14 - 13 = 1.
pOH₂ = 14 - 12 = 2.
[OH⁻]₁ = 10∧(-pOH₁) = 10⁻¹ M = 0.1 M.
[OH⁻]₂ = 10(-pOH₂) = 10⁻² M = 0.01 M.
[OH⁻]₁ · V₁(NaOH) = [OH⁻]₂ · V₂(NaOH).
0.1 M · 0.1 L = 0.01 M · V₂(NaOH).
V₂(NaOH) = 1 L, we can add 900 mL of water to convert solution of sodium hydroxide from pH=12 to pH=13, or we can add some strong acid.
Eduard22sly

We can covert 100 mL of NaOH solution with a pH of 13 to a solution with a pH of 12 by adding 900 mL of water to the 100 mL of NaOH solution.

Determination of the pOH of NaOH solution with pH 13

  • pH = 13
  • pOH =?

pH + pOH = 14

13 + pOH = 14

Collect like terms

pOH = 14 – 13

pOH = 1

Determination of the molarity of NaOH solution with pH 13 (i.e pOH = 1)

We'll begin by calculating the concentration of the hydroxide ion [OH¯]

  • pOH = 1
  • Concentration of the hydroxide ion [OH¯] =?

pOH = –Log [OH¯]

1 = –Log [OH¯]

Multiply through by –1

–1 = Log [OH¯]

Take the antilog of –1

[OH¯] = Antilog (–1)

[OH¯] = 0.1 M

Finally, we shall determine the molarity of the NaOH solution.

NaOH(aq) —> Na⁺(aq) + OH¯(aq)

From the balanced equation above,

1 mole of NaOH contains 1 mole of OH¯.

Therefore,

0.1 M NaOH will also contain 0.1 M OH¯

Determination of the pOH of NaOH solution with pH 12

  • pH = 12
  • pOH =?

pH + pOH = 14

12 + pOH = 14

Collect like terms

pOH = 14 – 12

pOH = 2

Determination of the molarity of NaOH solution with pH 12 (i.e pOH = 2)

We'll begin by calculating the concentration of the hydroxide ion [OH¯]

  • pOH = 2
  • Concentration of the hydroxide ion [OH¯] =?

pOH = –Log [OH¯]

2 = –Log [OH¯]

Multiply through by –1

–2 = Log [OH¯]

Take the antilog of –2

[OH¯] = Antilog (–2)

[OH¯] = 0.01 M

Finally, we shall determine the molarity of the NaOH solution.

NaOH(aq) —> Na⁺(aq) + OH¯(aq)

From the balanced equation above,

1 mole of NaOH contains 1 mole of OH¯.

Therefore,

0.01 M NaOH will also contain 0.01 M OH¯

Determination of the volume of the diluted solution.

  • Molarity of stock (M₁) = 0.1 M
  • Volume of stock solution (V₁) = 100 mL
  • Molarity of diluted solution (M₂) = 0.01
  • Volume of diluted solution (V₂) =?

M₁V₁ = M₂V₂

0.1 × 100 = 0.01 × V₂

10 = 0.01 × V₂

Divide both side by 0.01

V₂ = 1000 mL

Determination of the volume of water needed

  • Volume of stock solution (V₁) = 100 mL
  • Volume of diluted solution (V₂) = 1000 mL
  • Volume of water needed =?

Volume of water = V₂ – V₁

Volume of water = 1000 – 100

Volume of water needed = 900 mL

Thus, to change the pH of 100 mL of NaOH solution from pH of 13 to pH of 12, added 900 mL of water to the initial solution.

Learn more about pH:

https://brainly.com/question/1195974

Learn more about dilution:

https://brainly.com/question/15022582

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