A is a solution containing 5.00 g of HNO3 in 500 cm\(^3) of solution. B is a solution of NaOH of unknown concentration. 21.30cm3 of A was titrate it with 25.0 cm3 portions of B using methyl orange as an indicator. calculate the: (i) concentration ot A In mol dm−3
(ii) concentration of B in mol dm−3.
(iii) concentration of B in gdm−3
molar mass HNO3 = approximately 63 g/mol.
molar mass NaOH = approximately 40 g/mol.
mols HNO3 = g/molar mass = 5.00/63 = 0.0794
1. Molarity HNO3 = moles/L = 0.0794/0.500 = 0.159 M = 0.159 moles/dm^3
2. HNO3 + NaOH ==> NaNO3 + H2O
mols HNO3 = M x L = 0.159 M x 0.0213 L = 0.00337
From the equation you see 1 mol HNO3 requires 1 mol NaOH; therefore, 0.00337 mols HNO3 will require 0.00339 mols NaOH. M NaOH = moles/L = 0.00339/0.025 L = 0.135 M = 0.135 mol/dm^3
3. 0.135 mol/dm^3 x 40 g/mol = ? g/dm^3
Yes
Well, this sounds like a great chemistry problem. Let's dive right in, shall we?
(i) To find the concentration of A in mol dm\(^{-3}\), we need to determine the number of moles of HNO\(_3\) in 21.30 cm\(^3\) of solution.
Now, HNO\(_3\) has a molar mass of approximately 63 g/mol. So, we can calculate the number of moles of HNO\(_3\) using the formula:
Number of moles = Mass of HNO\(_3\) / Molar mass of HNO\(_3\)
In this case, we have 5.00 g of HNO\(_3\), so the number of moles is:
Number of moles = 5.00 g / 63 g/mol
Now, we have to convert the volume from cm\(^3\) to dm\(^3\). Since 1 dm\(^3\) is equal to 1000 cm\(^3\), we have:
Volume = 21.30 cm\(^3\) / 1000 cm\(^3\) per dm\(^3\)
Finally, we can calculate the concentration of A in mol dm\(^{-3}\) using the formula:
Concentration = Number of moles / Volume
(ii) Moving on to the concentration of B in mol dm\(^{-3}\), we need to use the titration data. Since 25.0 cm\(^3\) of B was used to neutralize 21.30 cm\(^3\) of A, we can assume that the stoichiometric ratio between HNO\(_3\) and NaOH is 1:1. This means that the number of moles of NaOH used is the same as the number of moles of HNO\(_3\) present in solution A.
So, the concentration of B in mol dm\(^{-3}\) can be calculated using the same formula as before:
Concentration of B = Number of moles of NaOH / Volume of B
(iii) Lastly, to find the concentration of B in g dm\(^{-3}\), we need to convert the concentration from mol dm\(^{-3}\) to g dm\(^{-3}\). To do this, we use the molar mass of NaOH, which is approximately 40 g/mol. We multiply the concentration in mol dm\(^{-3}\) by the molar mass to get the concentration in g dm\(^{-3}\).
I hope that helps! Good luck with your chemistry calculations. Remember, always keep your solutions balanced...and your humor too!
To solve this problem, we'll use the principles of stoichiometry and titration to determine the concentrations of solutions A and B.
Step 1: Calculate the concentration of A in mol/dm³:
Given:
Mass of HNO₃ in A = 5.00 g
Volume of A = 500 cm³ = 0.5 dm³
The molar mass of HNO₃ (nitric acid) is 63.01 g/mol.
To calculate the concentration of A, we need to convert the mass of HNO₃ to moles and divide by the volume in liters.
Concentration of A (in mol/dm³) = (mass of HNO₃ / molar mass of HNO₃) / volume of A
Concentration of A = (5.00 g / 63.01 g/mol) / 0.5 dm³
Step 2: Calculate the concentration of B in mol/dm³:
Given:
Volume of B = 25.0 cm³ = 0.025 dm³ (per titration)
Volume of A used in titration = 21.30 cm³ = 0.02130 dm³
(Since 25.0 cm³ is used in each titration and 21.30 cm³ is the actual volume used)
To find the concentration of B, we will use the stoichiometry of the balanced equation between HNO₃ and NaOH. In this reaction, the ratio is 1:1, meaning one mole of HNO₃ reacts with one mole of NaOH.
Using the equation:
HNO₃ (aq) + NaOH (aq) → NaNO₃ (aq) + H₂O (l)
Concentration of A (in mol/dm³) * Volume of A (in dm³) = Concentration of B (in mol/dm³) * Volume of B (in dm³)
Concentration of B (in mol/dm³) = (Concentration of A * Volume of A) / Volume of B
Concentration of B = (Concentration of A * 0.02130 dm³) / 0.025 dm³
Step 3: Calculate the concentration of B in g/dm³:
To convert the concentration of B from mol/dm³ to g/dm³, we need to multiply by the molar mass of NaOH.
Concentration of B (in g/dm³) = Concentration of B (in mol/dm³) * molar mass of NaOH
Note: The molar mass of NaOH is 39.997 g/mol.
Now, plug in the values and calculate:
(i) Concentration of A = (5.00 g / 63.01 g/mol) / 0.5 dm³
(ii) Concentration of B = (Concentration of A * 0.02130 dm³) / 0.025 dm³
(iii) Concentration of B = Concentration of B (in mol/dm³) * 39.997 g/mol
To calculate the concentration of solution A in mol dm\(^{-3}\), we need to first determine the number of moles of HNO3 in the solution. We can use the formula:
\(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\)
The molar mass of HNO3 is 63 g/mol. Therefore, the number of moles of HNO3 in solution A is:
\(\text{moles of HNO3} = \frac{5.00 \, \text{g}}{63 \, \text{g/mol}}\)
To calculate the concentration of solution A, we will divide the number of moles by the volume in dm\(^3\) (500 cm\(^3\) = 0.5 dm\(^3\)):
\(\text{concentration of A} = \frac{\text{moles of HNO3}}{\text{volume of solution A}}\)
Now let's calculate the concentration of solution B in mol dm\(^{-3}\). Since we are titrating solution A with solution B, and using methyl orange as an indicator, we can assume that the reaction between HNO3 and NaOH is a 1:1 mole ratio:
\(\text{moles of HNO3} = \text{moles of NaOH}\)
Since the volume of solution B used for titration is 25.0 cm\(^3\), we can use the concentration of solution A (calculated in part (i)) to find the moles of NaOH:
\(\text{moles of NaOH} = \text{concentration of A} \times \text{volume of B}\)
Now that we have the number of moles of NaOH, we can calculate the concentration of solution B by dividing the moles by the volume in dm\(^3\) (25.0 cm\(^3\) = 0.025 dm\(^3\)):
\(\text{concentration of B} = \frac{\text{moles of NaOH}}{\text{volume of B}}\)
To calculate the concentration of solution B in g dm\(^{-3}\), we need to know the molar mass of NaOH, which is 40 g/mol. We can then multiply the concentration of B in mol dm\(^{-3}\) by the molar mass to get the concentration in g dm\(^{-3}\):
\(\text{concentration of B in g dm}^{-3} = \text{concentration of B} \times \text{molar mass of NaOH}\)