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

The pH of a solution can be calculated using the formula:

$ext{pH} = - ext{log}_{10}[ ext{H}^+]$

For a 1.0 × 10⁻⁵ mol/dm³ solution, we have:

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

To calculate the concentration, we need the concordant titration results, which are 22.13 cm³.

The total volume of sodium hydroxide used is:

ext{Moles of NaOH} = 0.105 imes rac{22.13}{1000}

Calculating this, we find:

$0.105 imes 0.02213 = 0.00232465 ext{ moles NaOH}$

Since the reaction ratio is 2:1 for NaOH to H₂SO₄, the moles of sulfuric acid are:

ext{Moles of H₂SO₄} = rac{0.00232465}{2} = 0.001162325

Now we find the concentration in mol/dm³ using the volume of the sulfuric acid (25.0 cm³):

ext{Concentration} = rac{0.001162325}{0.025} = 0.046493

Thus, the concentration is approximately 0.0465 mol/dm³.

A pipette measures a fixed volume accurately, which is essential for obtaining precise volumes of sulfuric acid. This is important in titrations to ensure consistency. A burette, on the other hand, allows for the measurement of variable volumes of sodium hydroxide. It can deliver titrants drop by drop, making it easier to accurately determine the endpoint of the titration.

First, we calculate the moles of sodium hydroxide using the formula:

$ext{Moles} = ext{Volume (dm³)} imes ext{Concentration (mol/dm³)}$

For 30 cm³, we convert to dm³:

$30 ext{ cm}^3 = 0.030 ext{ dm}^3$

Then calculate the moles:

$ext{Moles} = 0.030 imes 0.105 = 0.00315 ext{ mol}$

Now, using the molar mass of NaOH (40 g/mol), we find the mass:

$ext{Mass} = ext{Moles} imes ext{Molar mass} = 0.00315 imes 40 = 0.126 ext{ g}$