Final answer:
To neutralise a mixture of 0.1 M H2SO4 and 0.2 M HCl, we sum the moles of H+ ions produced by each acid then multiply by the molar mass of NaOH. We find that 4 grams of NaOH would be required for neutralisation.
Step-by-step explanation:
To calculate the amount in grams of NaOH required to neutralise an acid mixture of 0.1 M H2SO4 (100 mL) and 0.2 M HCl (400 mL), we first need to determine the number of moles of each acid.
For H2SO4 (which is a diprotic acid, meaning each molecule can donate two protons), moles can be calculated as follows:
Molarity (M) × Volume (L) = Moles
0.1 M × 0.1 L = 0.01 moles of H2SO4
Since it is diprotic, the number of moles of H+ ions = 0.01 moles × 2 = 0.02 moles.
For HCl (which is monoprotic), moles are:
0.2 M × 0.4 L = 0.08 moles of HCl
Therefore, the total moles of H+ ions from both acids equals 0.08 moles + 0.02 moles = 0.10 moles. NaOH is a monoprotic base, so one mole of NaOH will neutralise one mole of H+ ions.
The molar mass of NaOH is approximately 40 g/mol. To find the mass of NaOH needed:
Moles of NaOH × Molar mass of NaOH = Mass of NaOH
0.10 moles × 40 g/mol = 4 grams
Therefore, you would need 4 grams of NaOH to neutralise the mixture of acids.