K12Chemistry

Solutions, acids, and bases

Describe how substances dissolve, measure concentration, and reason carefully about acid-base behavior in water.

Goals and prerequisites

You will distinguish dissolution from reaction, compare solubility with dissolving rate, calculate molarity and dilution, classify acids and bases using proton transfer, estimate pH for strong aqueous solutions, and write a neutralization equation.

Prerequisites: Review ions, ionic bonding, and moles. Introductory pH calculations here assume dilute aqueous solutions, ideal behavior, and 25 °C unless stated otherwise.

Dissolving and solubility

A solution is a homogeneous mixture. The solvent does the dissolving; the solute is dispersed through it. When NaCl dissolves, water molecules surround separated Na+ and Cl− ions. No new compound is required for this physical process, and evaporating the water can leave solid NaCl again.

Solubility is the maximum amount that dissolves under stated conditions. Rate is how quickly dissolving happens. Stirring, smaller pieces, and warmer water often increase the rate for a solid, but they do not necessarily increase the equilibrium solubility. Solubility also depends on the substances, temperature, and for gases pressure. A saturated solution contains as much dissolved solute as those conditions allow.

Some solutes react while dissolving. An acid may transfer a proton to water, so “dissolves” and “reacts” should be checked at the particle level rather than treated as synonyms.

Concentration and dilution

Molarity, M, is moles of solute per liter of solution: M = n/V. Use the final solution volume, not just the solvent volume. Units are mol L−1, commonly written M.

Worked example: molarity

What is the molarity of 0.250 mol NaCl in 0.500 L solution? M = 0.250 mol ÷ 0.500 L = 0.500 mol L−1, or 0.500 M. The solution has 0.500 mol NaCl in every liter at that concentration.

Dilution adds solvent while conserving solute moles. Therefore M1V1 = M2V2, provided the volumes use the same units.

Worked example: dilution

How much 2.00 M stock is needed to make 250.0 mL of 0.100 M solution? V1 = (0.100 M × 250.0 mL) ÷ 2.00 M = 12.5 mL. Measure 12.5 mL stock and add water until the total solution volume is 250.0 mL. The water volume is not assumed to be exactly 237.5 mL because mixing volumes can change slightly.

Acids, bases, and strength

In the Brønsted-Lowry model, an acid donates a proton and a base accepts one. In water, H+ is associated with water molecules, so H3O+ is a useful representation: HCl + H2O → H3O+ + Cl−. Ammonia acts as a base: NH3 + H2O ⇌ NH4+ + OH−.

Acid strength describes how extensively an acid ionizes in water, linked to Ka. Concentration describes how much solute was placed in a volume. A dilute strong acid can have a lower hydronium concentration than a concentrated weak acid in a particular comparison. Never use “strong” as a synonym for “concentrated.”

A conjugate acid-base pair differs by one proton. In HCl/Cl−, HCl is the acid and Cl− its conjugate base.

pH, pOH, and temperature assumptions

For dilute aqueous solutions, pH = −log10[H3O+] and pOH = −log10[OH−]. At 25 °C, pH + pOH = 14.00 is a useful introductory relation because Kw = 1.00 × 10−14. The precise IUPAC definition uses hydrogen-ion activity rather than bare concentration, especially when solutions are not dilute.

Neutral means equal hydronium and hydroxide activities, not necessarily pH 7 at every temperature. At temperatures other than 25 °C, Kw changes, so the neutral pH changes. Also, pH can be outside the familiar 0 to 14 range in concentrated solutions; the simple concentration formulas then need qualification.

Worked example: pH of a strong acid

For 1.0 × 10−3 M HCl, assume complete dissociation and dilute ideal behavior. [H3O+] ≈ 1.0 × 10−3 M, so pH = −log(1.0 × 10−3) = 3.00. This result is an estimate under the stated assumptions, not a universal rule for every HCl solution.

Neutralization and titration reasoning

An acid-base neutralization transfers protons. For strong acid and strong base, the net ionic equation is H3O+(aq) + OH−(aq) → 2 H2O(l). Molecular equations include spectator ions, such as HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l). The word “neutralization” describes the reaction, while the final pH depends on amounts, acid/base strengths, and the solution conditions.

Worked example: neutralization amount

What volume of 0.200 M NaOH neutralizes 25.0 mL of 0.100 M HCl? Moles HCl = 0.0250 L × 0.100 mol L−1 = 0.00250 mol. The 1:1 equation requires 0.00250 mol NaOH. V = n/M = 0.00250 ÷ 0.200 = 0.0125 L = 12.5 mL.

Practice with explained answers

  1. Which dissolves faster: a sugar cube or the same mass of crushed sugar in stirred water? Does that prove different solubility?
    Answer and reasoning

    Crushed sugar generally dissolves faster because more surface area is exposed. This compares rate; it does not by itself show a different equilibrium solubility.

  2. Find M for 0.0300 mol solute in 0.150 L solution.
    Answer and reasoning

    M = 0.0300 ÷ 0.150 = 0.200 M.

  3. Find the final concentration when 20.0 mL of 1.50 M stock is diluted to 100.0 mL.
    Answer and reasoning

    M2 = (1.50 × 20.0) ÷ 100.0 = 0.300 M. Solute moles stayed constant.

  4. Which is stronger: 0.001 M HCl or 1.0 M acetic acid?
    Answer and reasoning

    HCl is the stronger acid because it ionizes more completely. Concentration alone cannot decide the hydronium concentration or strength comparison; the acetic acid Ka and actual equilibrium are needed.

  5. Estimate pOH at 25 °C for pH 4.00.
    Answer and reasoning

    pOH = 14.00 − 4.00 = 10.00 under the 25 °C aqueous assumption.

Common misconceptions

Water is not “used up” when a salt dissolves. A strong acid is not automatically concentrated, and a neutralization reaction does not guarantee pH 7 at all temperatures or for every weak acid/base pair. A pH number is logarithmic: a one-unit difference represents a tenfold difference in hydronium activity in the introductory model.

Sources for further study