A 5.0 M solution of HNO3 is titrated with 0.3 M NaOH. Identify the species that have the highest concenttrations in the solution being titrated halfway to the equivalence point.
A 25.15 ml of 0.35 m HNO3 was titrated with an unknown concentration of NaOH. The endpoint was reached when 29.35 ml of NaOh were delivered. Find the concentration of the titrant.
For my homework there are several questions like the ones above. If I understand how to do the ones above I think i can complete the other questions.
Well, chemistry can be a real titration sensation, huh? Let's dive right into it!
For the first question, when the solution is halfway to the equivalence point, the highest concentrations will be of the species that are left over. In this case, since HNO3 is being titrated with NaOH, the species with the highest concentration would be the one that remains after the reaction. So, at the halfway point, the highest concentrations would be of the remaining HNO3 and NaOH.
Now, onto the second question. We have 25.15 ml of 0.35 M HNO3 being titrated with an unknown concentration of NaOH until the endpoint is reached at 29.35 ml. To find the concentration of the titrant (NaOH), we can use the concept of molarity and the equation:
M1V1 = M2V2
Here, M1 represents the concentration of HNO3, V1 is the volume of HNO3 used, M2 is the concentration of NaOH, and V2 is the volume of NaOH used.
Substituting the given values:
(0.35 M)(25.15 ml) = M2(29.35 ml)
Now, you can solve for M2 by dividing both sides by 29.35 ml:
M2 = (0.35 M)(25.15 ml) / 29.35 ml
And there you have it, the concentration of the titrant (NaOH)!
I hope that helps you with your homework. Just remember, chemistry is all about balancing the equations and having a "moleful" of fun!
Certainly! I'm here to help you understand the process for solving these questions.
Question 1: A 5.0 M solution of HNO3 is titrated with 0.3 M NaOH. Identify the species that have the highest concentrations in the solution being titrated halfway to the equivalence point.
To determine the species with the highest concentrations halfway to the equivalence point, we need to know the stoichiometry of the reaction. In this case, the balanced equation is:
HNO3 + NaOH -> NaNO3 + H2O
The stoichiometric ratio between HNO3 and NaOH is 1:1. This means that for every mole of HNO3 added, one mole of NaOH is required to fully neutralize it.
Halfway to the equivalence point, exactly half of the HNO3 has been neutralized. Therefore, it means that half of the moles of NaOH have been added.
Since the stoichiometric ratio is 1:1, this also means that the concentration of HNO3 and NaOH will be the same halfway to the equivalence point. Therefore, the species with the highest concentration at this point will be HNO3.
Question 2: A 25.15 ml of 0.35 M HNO3 was titrated with an unknown concentration of NaOH. The endpoint was reached when 29.35 ml of NaOH were delivered. Find the concentration of the titrant.
To find the concentration of the NaOH titrant, we can use the concept of stoichiometry and the volume of each solution used.
First, let's write the balanced equation for the reaction:
HNO3 + NaOH -> NaNO3 + H2O
We can see that the stoichiometric ratio is 1:1, meaning that one mole of HNO3 reacts with one mole of NaOH.
Next, we calculate the number of moles of HNO3 used in the reaction:
moles of HNO3 = volume of HNO3 (in L) × concentration of HNO3 (in mol/L)
Given the volume of HNO3 is 25.15 mL (or 0.02515 L) and the concentration is 0.35 M, we have:
moles of HNO3 = 0.02515 L × 0.35 mol/L
Now, we need to determine the number of moles of NaOH used in the reaction. Since the stoichiometric ratio is 1:1, the number of moles of NaOH is equal to the number of moles of HNO3 used:
moles of NaOH = moles of HNO3
To find the concentration of the NaOH titrant, we divide the moles of NaOH by the volume of NaOH used (in L):
concentration of NaOH = moles of NaOH / volume of NaOH (in L)
Given the volume of NaOH used is 29.35 mL (or 0.02935 L), we have:
concentration of NaOH = moles of NaOH / 0.02935 L
Therefore, to find the concentration of the titrant, you need to calculate the moles of NaOH and substitute it into the above equation.
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For #2.
NaOH + HNO3 ==> NaNO3 + H2O
mols HNO3 = M x L = ?
mols NaOH = mols HNO3(from the coefficients in the balanced equation.)
M NaOH = mols NaOH/L NaOH
For #1.
Assume some volume of the 5.0 M HNO3.
Calculate the volume NaOH needed to titrate half of it. Calculate the NO3^-, Na^+, and H2O. Actually I don't think they internd for H2O to be one of the asnwers but technically I think it is.