# IB DP Chemistry: SL复习笔记1.2.8 Concentration Calculations

### Concentration Calculations

#### Step by step

• Concentration calculations involve bringing together the skills and knowledge you have acquired previously and applying them to problem solving
• You should be able to easily convert between moles, mass, concentrations and volumes ( of solutions and gases)
• The four steps involved in problem solving are:
• write the balanced equation for the reaction
• determine the mass/ moles/ concentration/ volume of the of the substance(s) you know about
• use the balanced equation to deduce the mole ratios of the substances present
• calculate the mass/ moles/ concentration/ volume of the of the unknown substance(s)

#### Worked Example

25.0 cm3 of 0.050 mol dm-3 sodium carbonate was completely neutralised by 20.0 cm3 of dilute hydrochloric acid.Calculate the concentration in mol dm-3 of the hydrochloric acid.

Step 1: Write the balanced equation for the reaction

Na2CO3  +  2HCl  →  2NaCl  +  H2O  +  CO2

Step 2: Determine the moles of the known substance, in this case sodium carbonate. Don't forget to divide the volume by 1000 to convert cm3 to dm3

moles = volume x concentration

amount (Na2CO3) = 0.0250 dm3 x 0.050 mol dm-3 = 0.00125 mol

Step 3: Use the balanced equation to deduce the mole ratio of sodium carbonate to hydrochloric acid:

1 mol of Na2CO3 reacts with 2 mol of HCl, so the mole ratio is 1 : 2

Therefore 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl

Step 4: Calculate the concentration of the unknown substance, hydrochloric acid

#### Worked Example

Calculate the volume of hydrochloric acid of concentration 1.0 mol dm-3 that is required to react completely with 2.5 g of calcium carbonate.

Step 1: Write the balanced equation for the reaction

CaCO3  +  2HCl  →  CaCl2  +  H2O  +  CO2

Step 2: Determine the moles of the known substance, calcium carbonate

Step 3: Use the balanced equation to deduce the mole ratio of calcium carbonate to hydrochloric acid:

1 mol of CaCO3 requires 2 mol of HCl

So 0.025 mol of CaCO3 requires 0.050 mol of HCl

Step 4: Calculate the volume of HCl required