Please help with this question...

Question

Can 0.18 HCl be used in the titration of 1,0 g of baking soda instead of 0.18 M H2SO4? If so, would the volume of HCl be the same or different? If different, how much more or less HCl solution would you need?

Answer 1

To answer this question, we need to use the balanced chemical equation for the reaction between baking soda (sodium bicarbonate) and hydrochloric acid:

NaHCO3 + HCl -> NaCl + CO2 + H2O

From this equation, we can see that the mole ratio between HCl and NaHCO3 is 1:1. This means that one mole of HCl is required to react with one mole of NaHCO3.

To determine if 0.18 M HCl can be used in the titration instead of 0.18 M H2SO4, we need to calculate the number of moles of NaHCO3 that react with 1.0 g of baking soda:

moles of NaHCO3 = mass / molar mass = 1.0 g / 84.01 g/mol = 0.0119 mol

Since the mole ratio between HCl and NaHCO3 is 1:1, we need 0.0119 moles of HCl to react with 1.0 g of baking soda.

Now we can use the formula:

moles = concentration x volume (in liters)

to find the volume of 0.18 M HCl needed to provide 0.0119 moles:

0.0119 mol = 0.18 M x volume (in liters)

volume (in liters) = 0.0119 mol / 0.18 M = 0.066 L

Finally, we can convert the volume to milliliters:

volume (in milliliters) = 0.066 L x 1000 mL/L = 66 mL

Therefore, 66 mL of 0.18 M HCl can be used in the titration of 1.0 g of baking soda instead of 0.18 M H2SO4. The volume of HCl needed is different from the volume of H2SO4 needed, which was not provided in the question.