Boric acid Solution Preparation

How many grams of boric acid to weigh for any molarity and volume, with a full recipe table.

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To prepare a solution of boric acid you need the grams, and grams is molarity × volume in litres × 61.831 g/mol. A litre of 1 M takes 61.831 g; 500 mL of 0.1 M takes 3.0916 g; 250 mL of 0.01 M takes 154.58 mg. Enter the strength you want and the volume you want it in, and the calculator returns the mass to weigh together with the same figure in milligrams and as a percentage strength.

The figure comes from the formula. Boric acid is 3 × 1.008 (H) + 10.81 (B) + 3 × 15.999 (O), which is 61.831 g/mol, taking standard atomic weights, and every page here computes that sum rather than quoting it. Oxygen makes up the largest share of the mass: 3 of the 7 atoms in a formula unit are oxygen, and they account for 47.997 g of the 61.831 g, or 77.63% by mass. The derivation table below breaks the whole molecule down element by element.

Order of operations matters more than precision here. Dissolve the solid in rather less than the final volume, then make up to the mark once it has all gone into solution and come back to room temperature — warm solutions read low when they cool. If the calculated mass is under about 20 mg, prepare ten times the strength and dilute it tenfold by pipette instead, because C₁V₁ = C₂V₂ carries less error than a marginal weighing.

An insecticide, a mild antiseptic and a flux for welding. Its saturated solution is only about 5% at room temperature. At 61.831 g/mol it is heavier than 3 of the 12 acids covered here, and that ranking matters more than it looks: acetic acid has a molar mass of 60.052 g/mol, so a gram of it contains 2.96% more formula units than a gram of boric acid. Weigh by mass and you are not weighing equal amounts of substance.

One more step applies if you are starting from a concentrated bottle. Concentrated acids are labelled as a percentage by weight together with a density, not as a molarity, so getting to mol/L means multiplying density by that percentage and dividing by 61.831 g/mol. A page like this one gives you the last of those three numbers; the first two are on the bottle, and they vary between grades.

The formula

m = c × (V ÷ 1000) × 61.831
c
The molarity you want, in mol/L
V
The volume you want to make, in millilitres
61.831
The molar mass of boric acid in g/mol, derived from its formula
m
The mass of boric acid to weigh out, in grams

How it works, step by step

  1. Enter the molarity you want, in mol/L.
  2. Enter the volume you want to make, in millilitres.
  3. The two are multiplied together and by 61.831 g/mol.
  4. The result is the mass to weigh; dissolve it in part of the volume and then make up to the mark.

Worked examples

250 mL of 0.1 M boric acid

Weigh 1.54578 g. That is 0.1 M × 0.25 L × 61.831 g/mol, and it comes to 1545.78 mg if your balance is set to milligrams. Dissolve in about 175 mL first, then make up to 250 mL.

500 mL of 0.5 M boric acid

Weigh 15.4578 g. That is 0.5 M × 0.5 L × 61.831 g/mol, and it comes to 15,457.8 mg if your balance is set to milligrams. Dissolve in about 350 mL first, then make up to 500 mL.

1000 mL of 1 M boric acid

Weigh 61.831 g. That is 1 M × 1 L × 61.831 g/mol, and it comes to 61,831 mg if your balance is set to milligrams. Dissolve in about 700 mL first, then make up to 1000 mL.

How to read your score

0–0NothingNothing to weigh.
0–0.05Below a sensible weighingUnder 50 mg. Weigh ten or a hundred times this and dilute; the balance error does not shrink with the sample.
0.05–10Weighable directlyFrom 50 mg to 10 g, which is where a two- or three-decimal balance gives better than 1% on the mass.
10–—Large batchMore than 10 g of boric acid. Worth checking solubility and, for a solid that generates heat on dissolving, adding it in portions.

Frequently asked questions

What is the molar mass of boric acid (H3BO3)?

It is 61.831 g/mol. That is the sum of the standard atomic weights of every atom in the formula: 3 × 1.008 (H) + 10.81 (B) + 3 × 15.999 (O). One mole of boric acid therefore weighs 61.831 g, and a gram of it is 16.173 mmol.

What percentage of boric acid is oxygen?

77.63% by mass. Each formula unit contains 3 oxygen atoms contributing 47.997 g of the 61.831 g total, so 100 g of boric acid contains 77.63 g of oxygen and a kilogram contains 776.26 g of it.

How much boric acid do I need for 1 litre of 1 M solution?

61.831 g — the molar mass in grams, which is what a 1 molar solution means. For 1 L of 0.1 M it is 6.1831 g, and for 1 L of 0.01 M it is 0.61831 g.

How much for 100 mL of 0.1 M?

0.61831 g. The volume is a tenth of a litre and the strength a tenth of molar, so the mass is a hundredth of 61.831 g. In milligrams that is 618.31 mg.

Do I dissolve the solid first or make up the volume first?

Dissolve first, in perhaps 70% of the final volume, then make up to the mark. Adding solid to a full flask overshoots the volume and leaves the solution weak, and undissolved solid at the mark means the strength keeps changing as it goes in.

Can I dilute a stock solution instead of weighing?

Yes, and it is more accurate for small amounts. C₁V₁ = C₂V₂: to get 500 mL of 0.01 M from a 0.5 M stock, take 10 mL of stock and make up to 500 mL. Weighing the 309.16 mg that 500 mL of 0.01 M needs directly is the harder of the two.

Solution recipes for boric acid

How the molar mass of boric acid is arrived at
ElementAtomsAtomic weightContribution (g/mol)By mass
Oxygen (O)315.99947.99777.63%
Boron (B)110.8110.8117.48%
Hydrogen (H)31.0083.0244.891%
Total — one mole of boric acid61.831100%

Standard atomic weights, IUPAC 2021. The contribution column is atoms × atomic weight, and the total is the molar mass this page uses: 61.831 g/mol.

Grams of boric acid needed for a standard solution
Targetfor 100 mLfor 250 mLfor 500 mLfor 1 L
0.001 M6.1831 mg15.4578 mg30.9155 mg61.831 mg
0.005 M30.9155 mg77.2888 mg154.578 mg309.155 mg
0.01 M61.831 mg154.578 mg309.155 mg618.31 mg
0.05 M309.155 mg772.888 mg1.54578 g3.09155 g
0.1 M618.31 mg1.54578 g3.09155 g6.1831 g
0.15 M927.465 mg2.31866 g4.63732 g9.27465 g
0.2 M1.23662 g3.09155 g6.1831 g12.3662 g
0.25 M1.54578 g3.86444 g7.72888 g15.4578 g
0.5 M3.09155 g7.72888 g15.4578 g30.9155 g
1 M6.1831 g15.4578 g30.9155 g61.831 g
2 M12.3662 g30.9155 g61.831 g123.662 g

Each cell is molarity × volume in litres × 61.831 g/mol. Dissolve the solid first and then make up to the marked volume — adding solid to a full volume overshoots it.

Boric acid among other acids, by molar mass
CompoundFormulaMolar mass (g/mol)Millimoles in 1 g
Hydrochloric acidHCl36.45827.429
Formic acidCH2O246.02521.727
Acetic acidCH3COOH60.05216.652
Boric acid (this page)H3BO361.83116.173
Nitric acidHNO363.01215.87
Oxalic acidC2H2O490.03411.107
Lactic acidC3H6O390.07811.101

Ordered by molar mass. The last column is 1000/M, which is the number a weighed gram actually gives you.

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