Iodic acid partially dissociates into H+ and IO3-
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will be released as that of H+, its concentration is also X. The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3];
<span>Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; </span>
<span>At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; </span>
<span>so 0.17 = x² / (0.20 - x); </span>
<span>Solving for x using the quadratic formula: </span>
<span>x = [H+] = 0.063 M or pH = - log [H+] = 1.2.</span>
Hello!
A buffer is composed of a weak acid and its conjugate base or a weak base and its conjugate acid. From the given list:
HCl and HF: Strong Acid and Weak Acid. NOT BUFFER
HF and NaF: Weak Acid and Conjugate Base. BUFFER
HC₂H₃O₂ and KC₂H₃O₂: Weak Acid and Conjugate Base. BUFFER
NaOH and NH₃: Strong Base and Weak Base. NOT BUFFER.
Have a nice day!
A positively charged nucleus with two protons and two neutrons hope this helps :)
Answer:
And that means fish must be able to survive down here. This is due to a special property of water: the elasticity of H2O. We know that when the temperature sinks below freezing, water first contracts and then expands as it begins to turn to ice. Ice, being lighter than water, floats.
Explanation:
The property of water that allows fish to survive in a lake that is frozen over is the fact that ice is less dense than liquid water.
