A buret is partially filled with NaOH solution, to a volume of 1.14mL. A 20-mL sample of 0.1011 M HCl is titrated to a faint pink phenolphthalein endpoint. The final buret buret reading is 22.37mL. What is the molarity of the NaOH solution? Please help. Thank you in advance.
NaOH + HCl ==> NaCl + H2O
mols HCl = M x L = ?
Since the equation shows 1 mol NaOH = 1 mol HCl then mols NaOH = mols HCl.
Then M NaOH = mols NaOH/L NaOH.
You know mols NaOH and you know L NaOH. The only complicating factor is that the buret was not full for the titration. mL NaOH added = 22.37-1.14 = ? Convert that to L and substitute into the above. Solve for M NaOH.
Great! Thank you. This helps a lot!
penn sate
To calculate the molarity of the NaOH solution, you need to use the concept of stoichiometry and the balanced chemical equation for the reaction between NaOH and HCl.
The balanced chemical equation for the reaction is:
NaOH + HCl -> NaCl + H2O
From the equation, you can see that the stoichiometric ratio between NaOH and HCl is 1:1. This means that 1 mole of NaOH reacts with 1 mole of HCl.
Now let's calculate the number of moles of HCl used in the titration:
Moles of HCl = concentration of HCl x volume of HCl used
= 0.1011 M x 0.020 L
= 0.002022 moles
Since the stoichiometric ratio between NaOH and HCl is 1:1, the number of moles of NaOH is also 0.002022 moles.
Now, let's calculate the volume of NaOH used in the titration:
Volume of NaOH used = final buret reading - initial buret reading
= 22.37 mL - 1.14 mL
= 21.23 mL
Convert the volume of NaOH used to liters:
Volume of NaOH used = 21.23 mL x (1 L / 1000 mL)
= 0.02123 L
Finally, to calculate the molarity of the NaOH solution, divide the moles of NaOH by the volume of NaOH used:
Molarity of NaOH solution = moles of NaOH / volume of NaOH used
= 0.002022 moles / 0.02123 L
≈ 0.095 M
Therefore, the molarity of the NaOH solution is approximately 0.095 M.