how much .10M Base (in mL) is needed to neutralize 8.00mL of .10M Acid?
base- NaOH, acid- HCl, base volume (mL)- ?
base- NaOH, acid- HC2H3O2, base volume (mL)- ?
VaNa=VbNb
Vb= 8ml*.10N/.10N= 8m.
To determine how much 0.10M NaOH (base) is needed to neutralize 8.00mL of 0.10M HCl (acid), you can use the formula VaNa = VbNb, where Va is the volume of acid, Na is the normality of acid, Vb is the volume of base, and Nb is the normality of base.
Given:
Va = 8.00 mL
Na = 0.10M
Since Vb is what we need to find, we rearrange the equation to solve for Vb:
Vb = Va(Na/Nb)
Since both the normality of the acid and base are 0.10N, we can simplify the equation to:
Vb = 8.00 mL(0.10N/0.10N)
Vb = 8.00 mL
Therefore, 8.00 mL of 0.10M NaOH (base) is needed to neutralize 8.00 mL of 0.10M HCl (acid).
If the acid is changed to HC2H3O2, you would follow the same calculation, but substitute the normality of the acid (Nb) as 0.10N.