Math Problem Statement

45 cm³ of a hydrochloric acid solution of concentration of 0.2 mol·dm⁻³ is neutralised with a potassium hydroxide solution. 35 cm³ of the potassium hydroxide solution is required for neutralisation to take place. What is the concentration of the potassium hydroxide solution?

Solution

To solve this problem, we will use the concept of molarity and the stoichiometry of the neutralization reaction given.

Given Data:

  1. Volume of hydrochloric acid (HClHCl) solution, VHCl=45cm3V_{\text{HCl}} = 45 \, \text{cm}^3.
  2. Concentration of hydrochloric acid solution, CHCl=0.2moldm3C_{\text{HCl}} = 0.2 \, \text{mol} \, \text{dm}^{-3}.
  3. Volume of potassium hydroxide (KOHKOH) solution required for neutralization, VKOH=35cm3V_{\text{KOH}} = 35 \, \text{cm}^3.
  4. The reaction is: HCl+KOHKCl+H2OHCl + KOH \rightarrow KCl + H_2O

Steps to Solve:

  1. Convert Volumes to dm3\text{dm}^3: VHCl=45cm3=0.045dm3V_{\text{HCl}} = 45 \, \text{cm}^3 = 0.045 \, \text{dm}^3 VKOH=35cm3=0.035dm3V_{\text{KOH}} = 35 \, \text{cm}^3 = 0.035 \, \text{dm}^3

  2. Calculate the Moles of HClHCl: Using the formula Moles=Concentration×Volume\text{Moles} = \text{Concentration} \times \text{Volume}: Moles of HCl=CHCl×VHCl=0.2moldm3×0.045dm3\text{Moles of } HCl = C_{\text{HCl}} \times V_{\text{HCl}} = 0.2 \, \text{mol} \, \text{dm}^{-3} \times 0.045 \, \text{dm}^3 =0.009moles= 0.009 \, \text{moles}

  3. Using the Stoichiometry of the Reaction: The balanced equation shows a 1:1 molar ratio between HClHCl and KOHKOH. Therefore, moles of KOHKOH required for neutralization will also be 0.0090.009 moles.

  4. Calculate the Concentration of KOHKOH: Using Concentration=MolesVolume\text{Concentration} = \frac{\text{Moles}}{\text{Volume}}: CKOH=Moles of KOHVKOH=0.009moles0.035dm3C_{\text{KOH}} = \frac{\text{Moles of } KOH}{V_{\text{KOH}}} = \frac{0.009 \, \text{moles}}{0.035 \, \text{dm}^3} =0.257moldm3= 0.257 \, \text{mol} \, \text{dm}^{-3}

Answer:

The concentration of the potassium hydroxide solution is approximately 0.257moldm30.257 \, \text{mol} \, \text{dm}^{-3}.

Would you like more details or have any questions?

Here are some related questions for further practice:

  1. How would the concentration change if the volume of potassium hydroxide used was doubled?
  2. What volume of 0.1 mol/dm³ KOHKOH would be needed to neutralize the same amount of HClHCl?
  3. How would you adjust the calculation if the HClHCl solution was diluted before use?
  4. What if the reaction involved sulfuric acid (H2SO4H_2SO_4) instead of HClHCl; how would the stoichiometry change?
  5. How would the solution's molarity be affected if the temperature changes?

Tip: Always ensure units are consistent throughout your calculations to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Stoichiometry
Molarity
Unit Conversion

Formulas

Moles = Concentration × Volume
Concentration = Moles / Volume

Theorems

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Suitable Grade Level

Grade 10-12