Answer:
pH of the buffer is 7.33
Explanation:
The mixture of the ions H₂PO₄⁻ and HPO₄²⁻ produce a buffer (The mixture of a weak acid, H₂PO₄⁻, with its conjugate base, HPO₄²⁻).
To find pH of a buffer we use H-H equation:
pH = pka + log [A⁻] / [HA]
<em>Where A⁻ is conjugate base and HA weak acid.</em>
<em />
For the H₂PO₄⁻ and HPO₄²⁻ buffer:
pH = pka + log [HPO₄²⁻] / [H₂PO₄⁻]
Computing values of the problem:
pH =7.21 + log [0.125M] / [0.095M]
pH = 7.33
<h3>pH of the buffer is 7.33</h3>
<em />
If you have to write<span> the chemical </span>formula<span> of a simple, binary ionic compound given the name of the compound, you follow a set of three steps. Let's go through them using magnesium chloride as an example. </span>Write<span> the symbols for the cation and the anion: Mg and Cl. Determine the charge on the cation and anion.</span>
Answer:
k = 2,04x10⁻⁵
Explanation:
The equilibrium of acetic acid (CH₃COOH) in water is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺.
And the equilibrium constant is defined as:
k = [CH₃COO⁻] [H⁺] / [CH₃COOH] <em>(1)</em>
The equiibrium concentration of each specie if the solution of acetic acid is 0,05M is:
[CH₃COOH] = 0,05M - x
[CH₃COO⁻] = x
[H⁺] = x
<em>-Where x is the degree of reaction progress-</em>
As the pH is 3, [H⁺] = 1x10⁻³M. That means x = 1x10⁻³M
Replacing in (1):
k = (1x10⁻³)² / 0,05 - 1x10⁻³
k = 1x10⁻⁶ / 0,049
<em>k = 2,04x10⁻⁵</em>
<em></em>
I hope it helps!
It should be 2+ because Ba is in the 2nd group which means it needs two more electrons.
