To determine the molarity of the sodium hydroxide (NaOH) solution, you need to use the data from the titrations.
The equation for the neutralization reaction between NaOH and HCl is:
NaOH + HCl -> NaCl + H2O
Given that the volume of HCl solution used in each titration is 10 mL and the molarity of HCl is 0.28 M, we can calculate the number of moles of HCl used in each titration as follows:
moles of HCl = volume (in L) x molarity
moles of HCl = 0.01 L x 0.28 mol/L
moles of HCl = 0.0028 mol
Since the balanced equation shows a 1:1 ratio between NaOH and HCl, the same number of moles of NaOH is used in the neutralization reaction. Therefore, to find the molarity of NaOH, we need to divide the moles of NaOH by the volume of NaOH used in each titration.
Molarity of NaOH = moles of NaOH / volume (in L) of NaOH
Calculating the molarity for each titration:
1) Molarity of NaOH = 0.0028 mol / 0.00803 L = 0.348 M
2) Molarity of NaOH = 0.0028 mol / 0.00818 L = 0.342 M
3) Molarity of NaOH = 0.0028 mol / 0.00794 L = 0.352 M
To find the average molarity of NaOH, you can add up the individual molarities and divide by the number of titrations:
Average Molarity of NaOH = (0.348 M + 0.342 M + 0.352 M) / 3
Average Molarity of NaOH = 0.347 M
Therefore, the molarity of the sodium hydroxide solution is approximately 0.347 M.