This question is about acids and alkalis - AQA - GCSE Chemistry - Question 9 - 2018 - Paper 1

To find the pH of a 1.0 × 10⁻⁵ mol/dm³ solution, we use the formula:

$ext{pH} = - ext{log} [ ext{H}^+]$

For hydrochloric acid, the concentration of hydrogen ions, [H⁺], is equal to the concentration of the acid:

$ext{pH} = - ext{log}(1.0 imes 10^{-5}) = 5.0$.

Using the concordant results (Titration 3 and 4): Average titre = (rac{22.10 + 22.15}{2} = 22.125 \text{ cm}^3 )

Convert the average titre to dm³: (22.125 ext{ cm}^3 = 0.022125 ext{ dm}^3)

Moles of NaOH used: ( moles = concentration imes volume = 0.105 imes 0.022125 = 0.002324375 ) moles

Since the reaction ratio is 2:1, moles of H₂SO₄ = ( \frac{0.002324375}{2} = 0.0011621875 ) moles

Now, calculate the concentration of H₂SO₄: $\text{Concentration} = \frac{moles}{volume} = \frac{0.0011621875}{0.025} = 0.0464875 \text{ mol/dm}^3$

Therefore, the concentration is approximately 0.0465 mol/dm³.

A pipette measures a fixed volume accurately, making it ideal for delivering the exact amount of dilute sulfuric acid required for the titration. On the other hand, a burette allows for the titration of sodium hydroxide to be done drop by drop, which is essential for determining the end point of the reaction accurately.

First, convert the volume from cm³ to dm³: (30.0 ext{ cm}^3 = 0.030 ext{ dm}^3)

Now calculate the moles of sodium hydroxide: ( moles = concentration \times volume = 0.105 \times 0.030 = 0.00315 ) moles

Then, using the molar mass of NaOH (40 g/mol):

( ext{mass} = moles \times molar \ mass = 0.00315 \times 40 = 0.126 ext{ g} ).

Thus, the mass of sodium hydroxide is 0.126 g.