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s2008m [1.1K]
3 years ago
14

Like all equilibrium constants, the value of Kw depends on temperature. At body temperature (37°C), Kw = 2.4 * 10-14. What are t

he [H3O+] and pH of pure water at body temperature?
Chemistry
1 answer:
Aleksandr [31]3 years ago
5 0

Answer:

pH = 6.8124

Explanation:

We know pH decreases with increase in temperature.

At room temperature i.e. 25⁰c pH of pure water is equal to 7

We know

Kw = [H⁺][OH⁻]...............(1)

where Kw = water dissociation constant

At equilibrium [H⁺] = [OH⁻]

So at 37⁰c i.e body temperature Kw = 2.4 × 10⁻¹⁴

From equation (1)

[H⁺]² = 2.4 × 10⁻¹⁴

[H⁺] = √2.4 × 10⁻¹⁴

[H⁺] = 1.54 × 10⁻⁷

pH  = - log[H⁺]

      = - log{1.54 × 10⁻⁷}

      = 6.812

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Calculate the pH of a solution in which [OH–] = 4.5 × 10–9M.
posledela

Answer:

5.65 is the pH.

Explanation:

I am assuming that you are asking for confirmation on your answer. The answer is 5.65.

You would do:

[pOH] = -log[OH-]

          = -log[4.5*10^-9]

         equals about 8.3468

To find pH your would subtract the pOH from 14.

14-8.3468 = 5.65 << Rounded to match the answer choices.

7 0
3 years ago
A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug
OverLord2011 [107]

Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Henderson-Hasselbalch formula to find pH in a buffer is:

pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Replacing with known values:

5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

0.260 =  log₁₀ [CH₃COO⁻] / [CH₃COOH]

1.820 = [CH₃COO⁻] / [CH₃COOH] <em>(1)</em>

As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M <em>(2)</em>

Replacing (2) in (1):

1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

[CH₃COOH] = 0.100M / 2.820

[CH₃COOH] = <em>0.035M</em>

Thus: [CH₃COO⁻] = 0.100M - 0.035M = <em>0.065M</em>

5.40 mL of a 0.490 M HCl are:

0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = <em>7.429x10⁻³ moles of CH₃COO⁻</em>

<em />

Moles of CH₃COOH  before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles +  2.646x10⁻³ moles = <em>8.071x10⁻³ moles of CH₃COOH</em>. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

pH = 4.704

As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = <em>0.296</em>

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