GCSE Chemistry (AQA)
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Calculating Concentration of a Reacting Solution
- When two solutions react completely, they react in the ratio found in the balanced equation.
- Example: Potassium hydroxide (KOH) and hydrochloric acid (HCl) react in a 1:1 ratio.
- Balanced equation: 1 KOH + 1 HCl ��� 1 KCl + 1 H2O.
- Another example: Potassium hydroxide (KOH) and sulfuric acid (H2SO4) react in a 2:1 ratio.
- Balanced equation: 2 KOH + H2SO4 ��� K2SO4 + 2 H2O.
- The ratio of reactants can be used to calculate unknown concentrations.
- Example: Using titration to find the exact amount of KOH that neutralizes a known volume of HCl.
- If the concentration of HCl is known, the concentration of KOH can be calculated.
- Example data: 0.02 dm�� of KOH reacts with 0.05 dm�� of HCl with a concentration of 0.5 mol/dm��.
- Use the equation: amount of solute (moles) = concentration (mol/dm��) �� volume (dm��).
- Calculation: 0.05 dm�� HCl �� 0.5 mol/dm�� = 0.025 moles HCl.
- Since the ratio is 1:1, there are also 0.025 moles of KOH.
- Calculate the concentration of KOH: concentration (mol/dm��) = amount of solute (moles) / volume (dm��).
- Calculation: 0.025 moles / 0.02 dm�� = 1.25 mol/dm��.
- Example problem: 20.0 cm�� of NaOH neutralizes 25.0 cm�� of 0.05 mol/dm�� H2SO4.
- Balanced equation: 2 NaOH + H2SO4 ��� Na2SO4 + 2 H2O.
- Convert volumes to dm��: 20.0 cm�� = 0.02 dm��, 25.0 cm�� = 0.025 dm��.
- Calculate moles of H2SO4: 0.05 mol/dm�� �� 0.025 dm�� = 0.00125 moles.
- Use the ratio 2:1 to find moles of NaOH: 0.00125 moles H2SO4 �� 2 = 0.0025 moles NaOH.
- Calculate concentration of NaOH: 0.0025 moles / 0.02 dm�� = 0.125 mol/dm��.
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