What volume of 5% vinegar solution is needed to neutralize 1.0g of baking soda if density is 1.01g/ml

To find the volume of 5% vinegar solution needed to neutralize 1.0g of baking soda, we need to consider the stoichiometry of the reaction.

The balanced chemical equation for the reaction between baking soda (NaHCO3) and vinegar (acetic acid, CH3COOH) is:
CH3COOH + NaHCO3 -> CH3COONa + H2O + CO2

From the equation, we can see that the ratio between acetic acid and baking soda is 1:1. This means that for every 1 mole of baking soda, we need 1 mole of vinegar.

First, we need to calculate the number of moles of baking soda in 1.0g. The molar mass of baking soda (NaHCO3) is:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O = 16.00 g/mol

Molar mass of NaHCO3 = 22.99 + 1.01 + (12.01 + 3(16.00)) = 84.01 g/mol

Number of moles of baking soda = mass / molar mass = 1.0g / 84.01 g/mol

Next, since the ratio of baking soda to vinegar is 1:1, the number of moles of vinegar needed is also 1.0 mol.

Now, we need to find the volume of the 5% vinegar solution that contains 1.0 mol of vinegar.

The density of the vinegar solution is given as 1.01 g/ml, which means that 1.01g of vinegar solution has a volume of 1.00 ml.

To find the volume of the vinegar solution needed, we need to convert the number of moles of vinegar to grams:
mass of vinegar = number of moles x molar mass

mass of vinegar = 1.0 mol x (12.01 g/mol + 1.01 g/mol + 16.00 g/mol) = 29.02 g

Finally, we can find the volume of the vinegar solution needed:
volume of vinegar solution = mass of vinegar / density of vinegar solution

volume of vinegar solution = 29.02 g / 1.01 g/ml

Therefore, the volume of 5% vinegar solution needed to neutralize 1.0g of baking soda is approximately 28.73 ml.