Final answer:
The question pertains to the ionization of a weak acid, which is 3.0% dissociated in a 0.15 M solution. To find the concentration of hydrogen ions, we multiply the molar concentration of the acid (0.15 M) by the decimal fraction of dissociation (0.03), resulting in a hydrogen ion concentration of 0.0045 M.
Step-by-step explanation:
The student's question relates to the degree of ionization of a weak acid in solution. In Chemistry, when an acid is described as being a certain percentage dissociated, this refers to the extent to which the acid molecules separate into ions in a solution. Since the given weak acid is only 3.0% dissociated in a 0.15 M solution, we can use this percentage to calculate the concentration of hydrogen ions [H+] produced by the dissociation process. Remember that 3.0% equates to 0.030 in decimal form.
For instance, suppose we have acetic acid (CH3COOH), a typical weak acid. If we had a 1 M solution of acetic acid (concentration of acetic acid), only a small fraction, say 0.4%, would dissociate into ions (CH3COO- and H+), with the vast majority, 99.6%, remaining as undissociated CH3COOH molecules.
Thus, the dissociation of a weak acid in water is represented by an equilibrium expression, such as:
CH3CO₂H(aq) + H₂O(l) ⇒ H3O+ (aq) + CH3CO₂˅(aq)
With the given data, we can calculate the concentration of H+ ions generated by the 0.15 M weak acid solution when it's 3% dissociated. The calculation would be as follows:
[H+] = 0.15 M (concentration of the weak acid) x 0.03 (3% dissociation in decimal form) = 0.0045 M.
Therefore, the concentration of H+ in the solution is 0.0045 M.