25.0 ml of a 0.13M HCL solution is titrated with 17.6 ml of NaOH. What is the Molarity of the NaOH?
You only have 17.6ml of NaOH so it is 3.25/17.6 = 0.18
HCl + NaOH ==> NaCl + H2O
mols HCl = M x L = ?
mols NaOH = mols HCl (from the coefficients--they are 1:1)
Then M NaOH = mols NaOH/L NaOH.
M = ?
mols HCl= .13M x 25ml = 3.25 M ml
1:1 so mols NaOH = 3.25
M = (3.25 M ml /25 ml)
Correct
thanks
To find the molarity of NaOH solution, we need to use the equation for the balanced chemical reaction between HCl and NaOH:
HCl + NaOH -> NaCl + H2O
From the balanced equation, we can see that the mole ratio between HCl and NaOH is 1:1. This means that for every one mole of HCl, we require one mole of NaOH to completely react.
Given the volume and molarity of HCl solution, we can calculate the number of moles of HCl:
Moles of HCl = Volume of HCl solution (in liters) x Molarity of HCl solution
= 25.0 mL / 1000 mL/L x 0.13 mol/L
= 0.00325 moles of HCl
Since the mole ratio of HCl to NaOH is 1:1, the number of moles of NaOH is also 0.00325 moles.
Now, we can determine the molarity of NaOH solution using the equation:
Molarity of NaOH = Moles of NaOH / Volume of NaOH solution (in liters)
Moles of NaOH = 0.00325 moles
Volume of NaOH solution = 17.6 mL / 1000 mL/L = 0.0176 L
Molarity of NaOH = 0.00325 moles / 0.0176 L
= 0.1847 M
Therefore, the molarity of the NaOH solution is approximately 0.1847 M.