Solutions, Concentration & Acid–Base Basics

Solutions and the Language of Concentration

A solution is a homogeneous mixture in which the solute is dispersed uniformly throughout the solvent, the component present in greater amount. Aqueous solutions, in which water is the solvent, dominate general chemistry because so much chemistry — biological, environmental, and industrial — happens in water. Concentration expresses how much solute is packed into a given amount of solution, and choosing the right concentration unit (molarity, percent by mass, ppm, ppb) depends on the context and the sizes of the quantities involved.

• •Solute: substance present in smaller amount, being dissolved
• •Solvent: substance present in larger amount, doing the dissolving
• •Aqueous solution: denoted (aq), water is the solvent
• •Dilute vs concentrated describe relative, not absolute, amounts of solute

Molarity

Molarity (M) is defined as moles of solute per liter of solution and is the workhorse concentration unit in general chemistry because it connects directly to mole-based stoichiometry. Note carefully that the denominator is the volume of the total solution, not the volume of solvent added. Once you know molarity, it functions exactly like a conversion factor, letting you go from volume of solution to moles of solute and back.

• •M = mol solute / L solution
• •Moles solute = M × L solution
• •Volume solution = mol solute / M
• •Molarity changes with temperature slightly because volume expands, but moles do not

Preparing Solutions: Weighing and Dilution

There are two standard ways to prepare a solution of known molarity: weigh out a calculated mass of solid solute and dissolve it in solvent up to a defined final volume, or dilute a more concentrated stock solution with solvent. In dilution, only solvent is added — the moles of solute stay constant — so M₁V₁ = M₂V₂ relates the concentrated (1) and dilute (2) states. This equation is really just a statement that moles are conserved during dilution.

• •Weighing method: mass → moles (using molar mass) → dissolve to final volume in a volumetric flask
• •Dilution method: M₁V₁ = M₂V₂, add solvent only, moles solute unchanged
• •Volumetric flasks are used because they specify total solution volume precisely
• •Serial dilutions repeat the dilution equation in stages to reach very low concentrations

Solution Stoichiometry and Titration

Solution stoichiometry combines molarity with a balanced equation: convert given volume and molarity to moles, use mole ratios from the balanced equation, then convert back to whatever unit (moles, mass, volume, molarity) the question asks for. Titration is the classic application — a solution of known concentration (titrant) is added from a buret to a solution of unknown concentration until the reaction is exactly complete at the equivalence point, often signaled by a color-change indicator. At the equivalence point, moles of acid and moles of base are related in the exact stoichiometric ratio given by the balanced neutralization equation.

• •General path: mol A (given) → mol B (unknown) via mole ratio → concentration or volume of B
• •Equivalence point: stoichiometrically equal amounts of acid and base have reacted
• •Indicators change color near the equivalence point to signal the titration's end (endpoint)
• •For monoprotic acid + monobasic base: mol acid = mol base at equivalence

Electrolytes and What Dissolves in Water

When an ionic compound dissolves in water, it dissociates completely into its constituent ions, making the solution a strong electrolyte that conducts electricity well. Molecular compounds that ionize only partially (like weak acids and bases) are weak electrolytes, while molecular compounds that dissolve but do not ionize at all (like sugar) are nonelectrolytes. Correctly identifying which species are actually present in solution — molecules or ions — is essential for writing net ionic equations and understanding conductivity.

• •Strong electrolyte: 100% dissociation into ions (soluble ionic compounds, strong acids, strong bases)
• •Weak electrolyte: partial ionization, equilibrium mixture of molecules and ions (weak acids/bases)
• •Nonelectrolyte: dissolves as intact molecules, no ions formed (e.g., glucose, ethanol)
• •Solubility rules predict which ionic compounds dissolve appreciably in water

Solubility Rules for Ionic Compounds

Solubility rules are empirical generalizations that let you predict, without a solubility table, whether a given ionic compound will dissolve in water. Compounds containing alkali metal ions, ammonium, nitrate, and acetate are essentially always soluble, giving reliable starting points. Exceptions to otherwise-soluble or otherwise-insoluble categories (like Ag⁺, Pb²⁺, and Hg2²⁺ halides, or alkaline earth sulfates) must simply be memorized.

• •Always soluble: Group 1 cations, NH₄⁺, NO₃⁻, C₂H₃O₂⁻ (acetate), ClO₄⁻
• •Usually soluble: halides (Cl⁻, Br⁻, I⁻) except with Ag⁺, Pb²⁺, Hg2²⁺; sulfates except with Ba²⁺, Sr²⁺, Pb²⁺, Ca²⁺
• •Usually insoluble: carbonates, phosphates, hydroxides, sulfides, except with Group 1 or NH₄⁺
• •A precipitate forms when mixing two soluble solutions produces an insoluble combination of ions

Arrhenius and Brønsted–Lowry Acids and Bases

The Arrhenius definition, historically first, defines an acid as a substance that increases [H⁺] in water and a base as one that increases [OH⁻]. The more general Brønsted–Lowry definition defines an acid as a proton (H⁺) donor and a base as a proton acceptor, which explains acid–base behavior even without hydroxide ions, such as ammonia acting as a base. Every Brønsted–Lowry acid–base reaction produces a conjugate acid and a conjugate base, related to the original species by the gain or loss of a single proton.

• •Arrhenius acid: produces H⁺ (H₃O⁺) in water; Arrhenius base: produces OH⁻ in water
• •Brønsted–Lowry acid: proton donor; Brønsted–Lowry base: proton acceptor
• •Conjugate acid–base pair: differ by exactly one H⁺ (e.g., HA/A⁻, NH₄⁺/NH₃)
• •Water is amphiprotic — it can act as either an acid or a base depending on the partner

Strong Acids, Strong Bases, and Water's Autoionization

Only seven common strong acids ionize completely in water and must be memorized: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄, and HClO₃. Strong bases are the soluble hydroxides of Group 1 metals and the heavier Group 2 metals (Ca, Sr, Ba). Water itself undergoes a tiny autoionization, 2H₂O ⇌ H₃O⁺ + OH⁻, characterized at 25 °C by Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴, which holds in every aqueous solution, not just pure water.

• •Strong acids: HCl, HBr, HI, HNO₃, H₂SO₄ (first proton), HClO₄, HClO₃
• •Strong bases: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)₂, Sr(OH)₂, Ba(OH)₂
• •Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C
• •In pure water at 25 °C, [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ M

The pH Scale

pH = −log[H₃O⁺] compresses the huge range of possible hydronium concentrations into a convenient 0–14 scale at 25 °C, with pH + pOH = 14.00 always holding at that temperature. Because pH is a logarithm, the number of digits after the decimal point in a pH value equals the number of significant figures in the original concentration — a common sig-fig trap. Solutions with pH < 7 are acidic, pH = 7 are neutral, and pH > 7 are basic, all defined relative to pure water at 25 °C.

• •pH = −log[H₃O⁺]; [H₃O⁺] = 10^(−pH)
• •pOH = −log[OH⁻]; [OH⁻] = 10^(−pOH)
• •pH + pOH = 14.00 at 25 °C
• •Sig figs in log values: digits after the decimal point = sig figs in the original number

Neutralization and Percent/ppm Concentrations

A neutralization reaction between a strong acid and a strong base produces water and a salt, and can be tracked with the same solution-stoichiometry tools used elsewhere. For very dilute solutions, chemists often prefer percent by mass, parts per million (ppm), or parts per billion (ppb) over molarity because these units better represent trace-level concentrations relevant to environmental and biological samples. All three are mass-ratio units multiplied by an appropriate power of ten (100, 10⁶, or 10⁹).

• •Neutralization: acid + base → salt + water (for strong acid/strong base)
• •Percent by mass = (mass solute / mass solution) × 100%
• •ppm = (mass solute / mass solution) × 10⁶; ppb = (mass solute / mass solution) × 10⁹
• •For dilute aqueous solutions, 1 ppm ≈ 1 mg solute per liter of solution

What is the molarity of a solution prepared by dissolving 14.6 g of NaOH (molar mass 40.00 g/mol) in enough water to make 250.0 mL of solution?

1. 1Convert mass to moles: 14.6 g ÷ 40.00 g/mol = 0.365 mol NaOH.
2. 2Convert volume to liters: 250.0 mL = 0.2500 L.
3. 3Apply M = mol / L: M = 0.365 mol / 0.2500 L.
4. 4Compute: M = 1.46 mol/L.

M = 1.46 M NaOH

How many milliliters of a 6.00 M HCl stock solution are needed to prepare 500.0 mL of 0.250 M HCl?

1. 1Identify M₁ = 6.00 M (stock), V₁ = unknown, M₂ = 0.250 M, V₂ = 500.0 mL.
2. 2Apply M₁V₁ = M₂V₂.
3. 3Solve for V₁: V₁ = (M₂V₂)/M₁ = (0.250 M × 500.0 mL)/6.00 M.
4. 4Compute: V₁ = 125 mL/6.00 = 20.8 mL.
5. 5Add water to this 20.8 mL of stock acid up to a total volume of 500.0 mL.

V₁ = 20.8 mL of 6.00 M HCl, diluted to 500.0 mL

What volume of 0.150 M NaOH is required to neutralize 25.00 mL of 0.100 M H₂SO₄? (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O)

1. 1Find moles of H₂SO₄: mol = M × V = 0.100 mol/L × 0.02500 L = 2.50 × 10⁻³ mol.
2. 2Use the mole ratio from the balanced equation: 2 mol NaOH per 1 mol H₂SO₄.
3. 3Moles NaOH needed = 2.50 × 10⁻³ mol × (2/1) = 5.00 × 10⁻³ mol.
4. 4Find volume of NaOH: V = mol / M = 5.00 × 10⁻³ mol / 0.150 mol/L.
5. 5Compute: V = 0.0333 L = 33.3 mL.

V = 33.3 mL of 0.150 M NaOH

Calculate the pH of a solution with [H₃O⁺] = 3.2 × 10⁻⁴ M.

1. 1Apply pH = −log[H₃O⁺].
2. 2pH = −log(3.2 × 10⁻⁴).
3. 3log(3.2 × 10⁻⁴) = −3.49.
4. 4pH = −(−3.49) = 3.49.
5. 5Check sig figs: 2 sig figs in concentration → 2 digits after decimal in pH.

pH = 3.49 (acidic)

A solution has pOH = 9.75. Find [H₃O⁺] and classify the solution as acidic, neutral, or basic.

1. 1Find pH: pH = 14.00 − pOH = 14.00 − 9.75 = 4.25.
2. 2Apply [H₃O⁺] = 10^(−pH) = 10^(−4.25).
3. 3Compute: [H₃O⁺] = 5.6 × 10⁻⁵ M.
4. 4Since pH = 4.25 is less than 7, the solution is acidic.

[H₃O⁺] = 5.6 × 10⁻⁵ M; solution is acidic

Q1. What is the molarity of a solution containing 2.00 mol of KCl dissolved in enough water to make 4.00 L of solution?

Q2. In the dilution equation M₁V₁ = M₂V₂, what physical quantity is being held constant?

Q3. Which of the following is classified as a strong electrolyte in water?

Q4. According to solubility rules, which compound is expected to be insoluble in water?

Q5. By the Brønsted–Lowry definition, a base is best described as a substance that:

Q6. What is the conjugate acid of NH₃?

Q7. Which of the following is NOT one of the seven common strong acids?

Q8. At 25 °C, if [H₃O⁺] = 1.0 × 10⁻² M, what is [OH⁻]?

Q9. A solution has pH = 9.20. This solution is:

Q10. How many significant figures are in [H₃O⁺] if the pH is reported as 4.37?

Q11. What are the products of the neutralization of HCl with NaOH?

Q12. A water sample contains 3 mg of lead per liter of solution. Expressed in ppm, this concentration is approximately:

Q13. Which volumetric technique is used to determine an unknown concentration by reacting it with a solution of known concentration to the equivalence point?

Q14. Which of the following correctly identifies the solvent in a solution made by dissolving 5 g of sugar in 200 g of water?