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>
H2SO4 or hydrogen sulfate is an acid and NaOH or sodium hydroxide is a base or an alkali. The reaction between an acid and a base or alkali produces a salt and water. The reaction between these substances is shown below:
H2SO4 (aq) + 2NaOH (aq)------>2H20 (L) + Na2SO4 (aq). The salt produced in this reaction is sodium sulfate.
Answer:
Precise and maybe accurate.
Explanation:
To know if it is accurate, determine whether or not your measurements are similar to the correct measurements
The answer is the principal quantum number
Hope this helps
