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aniked [119]
2 years ago
6

Can someone help me balance these chemical equations?

Chemistry
2 answers:
coldgirl [10]2 years ago
7 0
What equations ?
If you post them I can help!
nordsb [41]2 years ago
6 0
I can’t see the equations
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Draw conclusion: The density of water is 1.0 g/mL, or 1.0 g/cm². Look at the data in your table. How can you use the density of
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What is the pH of a 0.640 M solution of C₅H₅NHBr (Kb of C₅H₅N is 1.7 × 10⁻⁹)?
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The pH of a 0.64 M solution of pyridine (C₅H₅N) is 9.52.  

<h3>What is pH ?</h3>

A figure expressing the acidity or alkalinity of a solution on a logarithmic scale on which 7 is neutral, lower values are more acid and higher values more alkaline.

The equation for the protonation of the base pyridine is the following:

C₅H₅N + H₂O ⇄ C₅H₅NH⁺ + OH⁻   (1)

Kb = 1.7 × 10⁻⁹ (Given)

To calculate the pH of the solution we need to use the following equation:

pH + pOH = 14

<em>pH = 14 - pOH</em>

     =14 - [-log[OH⁻]]

    = 14 + log[OH⁻]

Now, we need to find the concentration of the OH⁻ ions. Since pyridine is a weak base, at the equilibrium we have (eq 1):

C₅H₅N  +  H₂O  ⇄  C₅H₅NH⁺  +  OH⁻

0.64 - x                          x              x

After entering the values of [C₅H₅N] = 0.64-x, [C₅H₅NH⁺] = x, and [OH⁻] = x, into equation (2) we can find the concentration of OH⁻:

1.7 × 10⁻⁹  =[C₅H₅NH⁺]  [OH⁻]  /  [C₅H₅N]

                = x . x / 0.64-x

1.7 × 10⁻⁹ (0.64-x) - x² = 0

Solving the above quadratic equation for x, we have :

  • x₁ = -3.32 x 10⁻⁵
  • x₂ = 3.32 x 10⁻⁵

Now, We can calculate the pH, after taking the positive value, x₂, (concentrations cannot be negative) and entering into above equation :

<em />

<em>pH = </em>14 + log[OH⁻]

     = 14 + log (3.32 x 10⁻⁵)

 

     = 9.52

Therefore, the pH of the solution of pyridine is 9.52.

Find more about pH here:

brainly.com/question/8834103?referrer=searchResults

#SPJ1

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