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iren2701 [21]
3 years ago
15

What is the pH of .0003 M of NaOH

Chemistry
1 answer:
s344n2d4d5 [400]3 years ago
8 0

We are given that the concentration of NaOH is 0.0003 M and are asked to calculate the pH

We know that NaOH dissociates by the following reaction:

NaOH → Na⁺ + OH⁻

Which means that one mole of NaOH produces one mole of OH⁻ ion, which is what we care about since the pH is affected only by the concentration of H⁺ and OH⁻ ions

Now that we know that one mole of NaOH produces one mole of OH⁻, 0.0003M NaOH will produce 0.0003M OH⁻

Concentration of OH⁻ (also written as [OH⁻]) = 3 * 10⁻⁴

<u>pOH of the solution:</u>

pOH = -log[OH⁻] = -log(3 * 10⁻⁴)

pOH = -0.477 + 4

pOH = 3.523

<u>pH of the solution:</u>

We know that the sum of pH and pOH of a solution is 14

pH + pOH = 14

pH + 3.523 = 14                              [subtracting 3.523 from both sides]

pH = 10.477                        

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1.
Leya [2.2K]

Answer:

V₂ = 530.5 mL

Explanation:

Given data:

Initial temperature = 20.0°C

Final temperature = 40.0 °C

Final volume = 585 mL

Initial volume = ?

Solution:

Initial temperature = 20.0°C (20+273 = 293 K)

Final temperature = 40.0 °C (40+273 = 323 K)

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₁ = V₂T₁ /T₂  

V₂ = 585 mL × 293 K / 323 K

V₂ = 171405 mL.K / 323 K

V₂ = 530.5 mL

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<h3>Further explanation</h3>

Given

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Required

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