What volume of 0.115M NaOH is required to reach the equivalence point in the titration of 25.00 mL of 0.200M of HF?
Ka(HF)=7.4 x 10^-4
Ka is not needed.
mL x M = mL x M.
To determine the volume of NaOH required to reach the equivalence point in the titration, we need to calculate the number of moles of HF initially present in the 25.00 mL solution.
First, let's convert the volume of HF solution from milliliters to liters:
25.00 mL = 25.00 mL * (1 L / 1000 mL) = 0.02500 L
Next, we can calculate the number of moles of HF using the given molarity:
moles of HF = volume (L) * molarity (mol/L)
moles of HF = 0.02500 L * 0.200 mol/L = 0.00500 mol
Since the molecule HF is a weak acid, it undergoes a reaction with NaOH in a 1:1 ratio. This means that 1 mole of HF reacts with 1 mole of NaOH.
By using the equation NaOH + HF -> NaF + H2O, we can determine that 1 mole of NaOH reacts with 1 mole of HF.
Since the concentration of NaOH (0.115 M) is given, we can use stoichiometry to calculate the volume of NaOH required to reach the equivalence point.
Using the equation moles = concentration * volume (in L), we can rearrange it to solve for volume:
volume = moles / concentration
So,
volume of NaOH = moles of HF / concentration of NaOH
volume of NaOH = 0.00500 mol / 0.115 mol/L
volume of NaOH = 0.04348 L or 43.48 mL
Therefore, approximately 43.48 mL of 0.115 M NaOH is required to reach the equivalence point in the titration.