Convert concentrations, work out how much to weigh, dilute, mix solutions and find the pH of acids, bases and buffers, with the steps shown and the molar mass taken from the formula.
A solution is described by a concentration, and there are many ways to state it. Molarity (mol/L) divides the amount of solute by the solution volume; molality (mol/kg) divides by the mass of solvent and does not change with temperature; mass fraction and % w/w are fractions by mass; % w/v and g/L are solute mass per volume; ppm and ppb are mass fractions of one part per million and per billion. Normality is molarity times the equivalence factor (H⁺ given per mole of acid, for example 2 for H2SO4 fully neutralized), and the tool asks for that factor instead of guessing it. Going from a mass-based unit to a volume-based one needs the solution density: commercial HCl at 37 % w/w with d = 1.19 g/mL has 37 × 10 × 1.19 ÷ 36.458 = 12.08 mol/L. The molar mass comes from the formula you type (hydrates such as CuSO4.5H2O, 249.68 g/mol, included) using the CIAAW 2024 atomic weights, or from rounded exam masses. To prepare a solution, m = C × V × M ÷ purity: 0.1 mol/L of NaCl in 1 L takes 5.844 g. Dilution follows C1V1 = C2V2, where V2 is the final volume: 41.4 mL of 12.08 mol/L HCl topped up to 500 mL gives 1 mol/L. In mixing, the amount of each solute is conserved and the final concentration is the sum of C × V divided by the total volume (additive volumes). The pH comes from a full charge balance that includes water autoionization (Kw = 1.0 × 10⁻¹⁴ at 25 °C): that is why 1 × 10⁻⁸ mol/L HCl gives pH 6.98, not 8. For weak acids and bases the tool shows the quadratic solution and, only for comparison, the √(K·c) approximation with the 5 % test; 0.1 mol/L acetic acid (Ka = 1.75 × 10⁻⁵) has pH 2.88. Buffers use Henderson-Hasselbalch with an exact check, and the titration curve marks the equivalence point (pH 8.73 for 0.1 mol/L acetic acid with 0.1 mol/L NaOH). All values are for 25 °C and ideal solutions: above 1 mol/L ion activity departs from concentration. This is not a tool for dosing pool products or medicinal solutions.
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Molarity is moles of solute per liter of solution: c = m ÷ (M × V). For 5.844 g of NaCl (58.44 g/mol) in 1 L of solution: 5.844 ÷ 58.44 = 0.1000 mol, so 0.1000 mol/L. V is the final solution volume, measured to the mark of the flask, not the volume of water used.
| Unit | Value | |
|---|---|---|
| Molarity (mol/L) | 0.100 mol/L | |
| Millimolarity (mmol/L) | 100 mmol/L | |
| Normality (eq/L) | 0.100 eq/L | |
| Molality (mol/kg) | 0.101 mol/kg | |
| Mole fraction (x) | 0.00181 | |
| Mass concentration (g/L) | 5.84 g/L | |
| Milligrams per liter (mg/L) | 5840 mg/L | |
| % mass/volume (g/100 mL) | 0.584 % m/v | |
| % volume/volume | needs solute density | |
| % mass/mass | 0.584 % m/m | |
| Mass fraction (titer) | 0.00584 | |
| Parts per million (ppm, mg/kg) | 5840 ppm | |
| Parts per billion (ppb, µg/kg) | 5.84 × 10⁶ ppb |
See also: Molar mass · Equation balancer · Rule of three
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