Complete the table below:
How much 0.10 M Base (in mL) is required to neutralize 8.00 mL of the 0.10 M acid?
Base Acid Base Volume (mL)
NaOH HCl
NaOH HC2H3O2
You need to know if the acid/base is 1:1 or not. If 1:1 you may use a simple
mL x M = mL x M
If not 1:1 you must go through moles and coefficients.
Both answers are 8. I don't know why or how, though.
To complete the table, we need to calculate the volume of NaOH (base) required to neutralize 8.00 mL of the 0.10 M acid for both hydrochloric acid (HCl) and acetic acid (HC2H3O2).
To find the volume of base required, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction. The equation for the reaction between a strong base (NaOH) and a strong acid (HCl) is:
NaOH + HCl → NaCl + H2O
From the equation, we can see that the mole ratio between NaOH and HCl is 1:1. This means that 1 mole of NaOH reacts with 1 mole of HCl.
Given that the acid concentration is 0.10 M, it means that there are 0.10 moles of HCl in 1 liter (1000 mL) of the solution. Therefore, in 8.00 mL of 0.10 M HCl, there are:
moles of HCl = (0.10 moles/L) * (8.00 mL / 1000 mL) = 0.0008 moles
Since the mole ratio between NaOH and HCl is 1:1, we need the same number of moles of NaOH to neutralize the acid. Hence, 0.0008 moles of NaOH is needed.
Now, to find the volume of NaOH in mL needed to achieve 0.0008 moles, we need to consider the concentration of NaOH (0.10 M).
moles of NaOH = moles of HCl = 0.0008 moles
Using the equation:
moles = concentration * volume (in liters)
volume (in liters) = moles / concentration
volume (in liters) = 0.0008 moles / 0.10 M = 0.008 L
Finally, we need to convert the volume from liters to milliliters since the question asks for the volume in mL:
volume (in mL) = 0.008 L * 1000 mL/L = 8.00 mL
Therefore, for both HCl and HC2H3O2, the volume of 0.10 M NaOH required to neutralize 8.00 mL of the acid would be 8.00 mL.