A 0.2512 stock sample of HCl is tirade to equivalence . If 16.77ml of 0.1410 M NaOH is required to reach the equivalence point, what volume of stock HCl was used for the tritation?
Is that 0.2512 M stock? I will assume it is.
mols NaOH = M x L = ?
mols HCl = mols NaOH (since 1 mol HCl = 1 mol NaOH in the balanced equation).
M HCL stock = mols/L.
You know M and you know mols, solve for L and change to mL if desired.
yes 0.2512 M stock so,
0.1410m x 0.01677L= 2.36457^-03 mole used.
mols HCL = mols NaOH 0.2512M=2.36457^3
What am I missing ?
Should there be another step?
To find the volume of the stock HCl used for the titration, we can use the equation:
(moles of HCl) = (moles of NaOH)
First, let's calculate the number of moles of NaOH used:
(moles of NaOH) = (volume of NaOH) * (concentration of NaOH)
We are given:
Volume of NaOH = 16.77 mL
Concentration of NaOH = 0.1410 M
Converting the volume to liters:
Volume of NaOH = 16.77 mL = 0.01677 L
(moles of NaOH) = (0.01677 L) * (0.1410 M)
(moles of NaOH) = 0.00236457 moles
Since HCl and NaOH react in a 1:1 ratio based on their balanced chemical equation, the number of moles of HCl is also 0.00236457 moles.
Now, let's calculate the volume of the stock HCl using its concentration:
(moles of HCl) = (volume of HCl) * (concentration of HCl)
We are given:
Concentration of HCl = 0.2512 M
(moles of HCl) = (volume of HCl) * (0.2512 M)
Rearranging the equation to solve for the volume of HCl:
Volume of HCl = (moles of HCl) / (0.2512 M)
Volume of HCl = 0.00236457 moles / (0.2512 M)
Volume of HCl = 0.009395 liters
Converting the volume to milliliters:
Volume of HCl = 0.009395 liters * 1000 mL/L
Volume of HCl = 9.395 mL
Therefore, 9.395 mL of the stock HCl was used for the titration.