A 21.00 sample of 0.3254M Hcl solution requires 26.09 ml of NaOH solution for complete neutralization.
Calculate the volume of NaOH solution in liters, required for the titration.
HCl + NaOH ==> NaCl + H2O
mols HCl = M x L = ?
mols NaOH = same. (Look at the coefficients in the balanced equation.)
M NaOH = mols NaOH/L NaOH. You have molos and M, solve for L.
To calculate the volume of NaOH solution required for the titration, we need to use the equation:
M₁V₁ = M₂V₂
Where:
M₁ is the molarity of the HCl solution (0.3254 M)
V₁ is the volume of the HCl solution (21.00 mL = 21.00/1000 L)
M₂ is the molarity of the NaOH solution (which we will calculate)
V₂ is the volume of the NaOH solution (which we need to find)
First, let's calculate the Moles of HCl:
Moles of HCl = Molarity * Volume
= 0.3254 M * (21.00/1000) L
Now, since the reaction between HCl and NaOH is 1:1, the moles of HCl and NaOH should be equal when the reaction is complete. So, the moles of NaOH can be calculated as follows:
Moles of NaOH = Moles of HCl
= 0.3254 M * (21.00/1000) L
Finally, let's find the volume of the NaOH solution:
Molarity of NaOH (M₂) = Moles of NaOH / Volume of NaOH (V₂)
We can now plug in the values:
M₂ * V₂ = 0.3254 M * (21.00/1000) L
V₂ = (0.3254 M * 21.00/1000) L / M₂
Now, we need the molarity (M₂) of the NaOH solution. However, the molarity is not given in the problem. We will need more information in order to calculate it.