In an experiment,20.0cm³ of a solution containing 4gdm-³ of sodium hydroxide was neutralized by 8.0cm³ of dilute tetraoxosulphate (vi) acid
1)Write the balanced equation
2) calculate the concentration of the acid in moldm³
1) NaOH + H2SO4 -> Na2SO4 + 2H2O
2) To calculate the concentration of the acid, we need to use the equation:
Moles of acid = Moles of sodium hydroxide
Moles of sodium hydroxide = volume x concentration
Moles of sodium hydroxide = 20.0cm³ x 4gdm-³ / 1000g/mol = 0.08 mol
Moles of acid = volume x concentration
0.08 mol = 8.0cm³ x concentration / 1000cm³/L
Concentration of acid = 10 mol/L or 10M
1) The balanced equation for the neutralization reaction between sodium hydroxide (NaOH) and tetraoxosulphate (VI) acid (H2SO4) is as follows:
2NaOH + H2SO4 → Na2SO4 + 2H2O
2) To calculate the concentration of the acid in moldm³ (moles per liter), we first need to find the number of moles of acid used.
Given that 8.0 cm³ of the acid was used, we can convert this volume to liters:
8.0 cm³ = 8.0 × 10^(-3) L (since 1 cm³ = 1 × 10^(-3) L)
Now, based on the balanced equation, we know that the mole ratio between NaOH and H2SO4 is 2:1. This means that for every 2 moles of NaOH, we need 1 mole of H2SO4.
Next, we need to find the number of moles of NaOH used.
The concentration of NaOH is given as 4 g dm^(-3). We can use the formula:
concentration (in moldm³) = mass (in grams) / molar mass (in g/mol)
To find the molar mass of NaOH, we need to calculate the sum of the atomic masses of each element in the compound:
Na (sodium) = 22.99 g/mol
O (oxygen) = 16.00 g/mol
H (hydrogen) = 1.01 g/mol
Molar mass of NaOH = (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol) = 40.00 g/mol
Now, we can calculate the number of moles of NaOH:
moles of NaOH = mass (in grams) / molar mass (in g/mol)
moles of NaOH = 4 g / 40.00 g/mol = 0.10 mol
Since the molar ratio is 2:1, the moles of H2SO4 used will be half of the moles of NaOH used:
moles of H2SO4 = 0.10 mol / 2 = 0.05 mol
Finally, we can calculate the concentration of the acid in moldm³.
concentration (in moldm³) = moles / volume (in liters)
concentration (in moldm³) = 0.05 mol / 8.0 × 10^(-3) L
concentration (in moldm³) = 6.25 moldm³