250ml of 0.1M HCl is titrated against 0.2M KOH. After 5ml of KOH was used up the rest neutralization was carried out with 0.02M .NaOH what was the volume of NaOH used?
how many moles HCl available?
how many moles KOH was used?
each mole of KOH neutralized 1 mole of HCl
Now, knowing how many moles of HCl are left, it will take that many moles of NaOH,
so convert that to ml of .02M NaOH.
To determine the volume of NaOH used, we can use the concept of stoichiometry and the equation of the reaction between HCl and NaOH:
HCl + NaOH -> NaCl + H2O
First, let's find the moles of HCl used. We know that the initial concentration is 0.1 M and the volume used is 250 ml.
Moles of HCl = concentration × volume
= 0.1 M × 0.250 L
= 0.025 mol
Since the reaction between HCl and KOH is 1:1, we can conclude that 0.025 mol of KOH was also used up.
Next, we need to calculate the amount of KOH remaining after the neutralization with 5 ml of 0.2 M KOH.
Moles of KOH neutralized with HCl = concentration × volume
= 0.2 M × 0.005 L
= 0.001 mol
Remaining moles of KOH = Initial moles of KOH - Moles of KOH neutralized with HCl
= 0.025 mol - 0.001 mol
= 0.024 mol
Now, let's find the volume of NaOH required to neutralize the remaining KOH. The concentration of NaOH is given as 0.02 M.
Volume of NaOH = Moles of KOH remaining / Concentration of NaOH
= 0.024 mol / 0.02 M
= 1.2 L
Therefore, the volume of NaOH used is 1.2 liters.
To find the volume of NaOH used, we can use the concept of stoichiometry and the balanced chemical equation of the reaction between HCl and NaOH.
First, let's write the balanced chemical equation for the reaction:
HCl + NaOH → NaCl + H2O
We know that the HCl solution has a concentration of 0.1M and a volume of 250ml, which means it contains 0.1 moles of HCl.
The KOH solution has a concentration of 0.2M and is titrated against the HCl solution. We are given that 5ml of KOH is used up, so we can calculate the number of moles of KOH used:
0.2 moles/liter × 0.005 liters = 0.001 moles of KOH
Since the reaction is 1:1 between HCl and KOH, this means that 0.001 moles of HCl have reacted as well.
Now, the remaining neutralization is carried out with NaOH, which has a concentration of 0.02M.
Using the stoichiometry of the balanced equation, we can determine that the mole ratio between NaOH and HCl is 1:1. This means that we need to use 0.001 moles of NaOH to react with the remaining 0.001 moles of HCl.
To find the volume of NaOH used, we can use the equation:
moles = concentration × volume
Rearranging the equation to solve for volume:
volume = moles / concentration
volume = 0.001 moles / 0.02 moles/liter
volume ≈ 0.05 liters
Since 1 liter is equal to 1000 ml, the volume of NaOH used is approximately 0.05 liters × 1000 ml/liter = 50 ml.