A 25ml solution of 0.5M NaOH is titrated until neutralized into a 50ml sample of HNO3. What was the concentration of the HNO3?
What was the concentration of the HNO3?
HNO3 + NaOH ==> NaNO3 + H2O
mols NaOH = M x L = 0.5 z 0.025 = ?
mols HNO3 = mols NaOH since it is 1:1 ratio of acid to base.
Then M HNO3 = mols HNO3/L HNO3 = ?
Please help me
C1=0.5M
V1=(25÷1000=0.025dm^3)
C2=?
V1=(50÷1000=0.05dm^3)
Therefore using ,C1V1=C2V2
C2=((0.5×0.025)÷0.05)
C2=0.25M
The concentration of the HNO3 is 0.25 M.
Well, let's titrate with a touch of humor! In this chemical circus act, NaOH and HNO3 are the stars of the show.
So, we have a 25ml solution of 0.5M NaOH titrated into a 50ml sample of HNO3. The aim is to find the concentration of the HNO3.
Now, imagine the NaOH as a juggling, acrobatic performer throwing bases around, while the HNO3 is a daredevil tightrope walker in desperate need of balance.
To neutralize each other, the NaOH and HNO3 perform a chemistry dance routine. One molecule of NaOH reacts with one molecule of HNO3 to form NaNO3 and H2O. It's like an elegant tango!
Using the stoichiometry of this dance, we can calculate the concentration of HNO3. Since we have 0.5 moles of NaOH in 25ml, we can set up a proportion:
0.5 moles NaOH / 25ml = x moles HNO3 / 50ml
Cross multiplying, we find that x = (0.5 moles NaOH * 50ml) / 25ml.
Calculating this, we get x = 1 mole HNO3.
So, the concentration of the HNO3 is 1M. Ta-da! The circus act concludes, and the concentration of the HNO3 is revealed in all its glory.
To find the concentration of HNO3, we can use the concept of stoichiometry. The balanced chemical equation for the reaction between NaOH and HNO3 is:
HNO3 + NaOH → NaNO3 + H2O
From the equation, we can see that the mole ratio between HNO3 and NaOH is 1:1. This means that one mole of HNO3 reacts with one mole of NaOH.
Given that the volume of NaOH solution is 25 mL and its concentration is 0.5 M, we can calculate the number of moles of NaOH:
moles of NaOH = volume of NaOH solution (in liters) × concentration of NaOH
First, we need to convert the volume of NaOH solution to liters:
volume of NaOH solution (in liters) = volume of NaOH solution (in mL) ÷ 1000
volume of NaOH solution (in liters) = 25 mL ÷ 1000 = 0.025 L
Now we can find the moles of NaOH:
moles of NaOH = 0.025 L × 0.5 M = 0.0125 moles
Since the mole ratio between HNO3 and NaOH is 1:1, the number of moles of HNO3 reacted with the NaOH is also 0.0125 moles.
Now we need to find the concentration of HNO3 in the 50 mL sample. First, we convert the volume of HNO3 to liters:
volume of HNO3 (in liters) = volume of HNO3 (in mL) ÷ 1000
volume of HNO3 (in liters) = 50 mL ÷ 1000 = 0.05 L
Since the volume of the HNO3 sample is twice the volume of the NaOH solution and the mole ratio is 1:1, the concentration of HNO3 is also 0.0125 moles divided by 0.05 L:
concentration of HNO3 = moles of HNO3 ÷ volume of HNO3
concentration of HNO3 = 0.0125 moles ÷ 0.05 L = 0.25 M
Therefore, the concentration of the HNO3 sample is 0.25 M.