25ml of 0.05M of NaOH and the average volume of HCl is 17.7ml after titration and the concentration of HCl is 0.1028M.
Find the [OH] eq (from titration).
How do u solve this...
To solve this problem, you can use the concept of stoichiometry to determine the concentration of hydroxide ions ([OH-]) in the NaOH solution.
First, you need to calculate the number of moles of HCl used in the titration. To do this, use the formula:
Moles of HCl = Volume of HCl solution (in L) × Concentration of HCl (in mol/L)
Given that the volume of HCl used in the titration is 17.7 mL and the concentration of HCl is 0.1028 M, convert the volume to liters:
Volume of HCl solution = 17.7 mL × (1 L / 1000 mL) = 0.0177 L
Now, calculate the moles of HCl:
Moles of HCl = 0.0177 L × 0.1028 mol/L = 0.00181656 mol
Since the reaction between HCl and NaOH is a 1:1 ratio, the moles of NaOH used in the titration is also 0.00181656 mol.
Now, you can determine the concentration of NaOH using the formula:
Concentration of NaOH (in mol/L) = Moles of NaOH / Volume of NaOH solution (in L)
Given that the volume of NaOH solution is 25 mL and the moles of NaOH is 0.00181656 mol, convert the volume to liters:
Volume of NaOH solution = 25 mL × (1 L / 1000 mL) = 0.025 L
Now, calculate the concentration of NaOH:
Concentration of NaOH = 0.00181656 mol / 0.025 L ≈ 0.07266 M
Since NaOH is a strong base and dissociates completely in water, the concentration of hydroxide ions ([OH-]) is equal to the concentration of NaOH.
Therefore, [OH-] eq (from titration) ≈ 0.07266 M.