1g of mixture na2c03 and k2c03 was made upto 250ml is aqueous solution. 25ml of this solution was neutralized 20ml of hcl of unknown concentration. The neutralized solution requires 16.24ml of 0.1N agn03 for precipitation Calculate A) the k2c03 is mixture b) conc of hcl in g/L c) molarity of hcl

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To solve this problem step-by-step, we'll first calculate the concentration of Na2CO3 and K2CO3 in the mixture:

Step 1: Calculate the number of moles of Na2CO3 and K2CO3 in the 25ml solution.

Molar mass of Na2CO3 = 22.99 (2) + 12.01 + 16.00 (3) = 105.99 g/mol

Molar mass of K2CO3 = 39.10 (2) + 12.01 + 16.00 (3) = 138.21 g/mol

Number of moles of Na2CO3 = mass / molar mass = 1g / 105.99 g/mol = 0.0094 mol

Number of moles of K2CO3 = mass / molar mass = 1g / 138.21 g/mol = 0.0072 mol

Step 2: Calculate the concentrations of Na2CO3 and K2CO3 in the 25ml solution.

Concentration (molarity) = number of moles / volume (in liters)

Concentration of Na2CO3 = 0.0094 mol / 0.025 L = 0.376 M

Concentration of K2CO3 = 0.0072 mol / 0.025 L = 0.288 M

Now, let's move on to part A) to calculate the concentration of K2CO3 in the mixture:

Step 3: Calculate the total volume of the mixture.

The solution was made up to 250ml, so the total volume is 250ml.

Step 4: Calculate the number of moles of K2CO3 in the mixture.

Number of moles of K2CO3 in the mixture = concentration of K2CO3 x total volume (in liters) = 0.288 M x 0.250 L = 0.072 mol

Therefore, the concentration of K2CO3 in the mixture is 0.072 mol/L.

Moving on to part B) to calculate the concentration of HCl in g/L:

Step 5: Calculate the number of moles of AgNO3 used in the titration.

Number of moles of AgNO3 = Molarity x Volume (in liters) = 0.1 mol/L x 0.01624 L = 0.001624 mol

Step 6: Calculate the number of moles of HCl neutralized by AgNO3.

The balanced chemical equation for the neutralization is:

2AgNO3 + 2HCl -> 2AgCl + 2HNO3

From the equation, we see that 2 moles of AgNO3 neutralize 2 moles of HCl.

Therefore, the number of moles of HCl neutralized = 0.001624 mol / 2 = 0.000812 mol

Step 7: Calculate the number of grams of HCl neutralized.

Molar mass of HCl = 1.01 (1) + 35.45 = 36.46 g/mol

Mass of HCl neutralized = number of moles x molar mass = 0.000812 mol x 36.46 g/mol = 0.0296 g

Step 8: Calculate the concentration of HCl in g/L.

Concentration (g/L) = mass / volume (in liters)

Concentration of HCl = 0.0296 g / 0.020 L = 1.48 g/L

Finally, let's move on to part C) to calculate the molarity of HCl:

Step 9: Calculate the molarity of HCl.

Molarity (M) = concentration (g/L) / molar mass (g/mol)

Molarity of HCl = 1.48 g/L / 36.46 g/mol = 0.040 M

Therefore, the molarity of HCl is 0.040 M.

To summarize our answers:

A) The concentration of K2CO3 in the mixture is 0.072 mol/L.
B) The concentration of HCl in g/L is 1.48 g/L.
C) The molarity of HCl is 0.040 M.

Sure, I can help you step by step!

To solve this problem, we will use the concept of stoichiometry and the equation balancing method. Here's how to approach each part of the question:

A) Calculate the amount of K2CO3 in the mixture:
1. First, let's determine the amount of Na2CO3 in the mixture.
- Given: 1g of the mixture.
- Na2CO3 has a molar mass of 106 g/mol, so 1g is equivalent to 1/106 moles.

2. Since the mixture is made up to 250ml, the concentration of the mixture is:
- Concentration = amount of solute / volume of solution.
- Concentration = (1/106) moles / 0.25 L = 4/425 moles per liter.

3. Now, since for every 1 mole of K2CO3 in the mixture, there are 2 moles of Na2CO3, we can find the amount of K2CO3.
- Amount of K2CO3 = (2/4) × (4/425) moles per liter.
- Amount of K2CO3 = 1/425 moles per liter.

B) Calculate the concentration of HCl in g/L:
1. From the given information, we know that 25ml of the solution neutralized 20ml of HCl.

2. To determine the concentration of HCl, we need to find the number of moles of HCl neutralized.
- Moles of HCl neutralized = Moles of AgNO3 added at the endpoint.
- Given: Volume of AgNO3 used = 16.24ml, Molarity of AgNO3 = 0.1N.
- Moles of AgNO3 = Molarity × Volume in liters.
- Moles of AgNO3 = 0.1 × (16.24/1000) moles.

3. The reaction between AgNO3 and HCl follows a 1:1 stoichiometric ratio, so the moles of HCl neutralized are also 0.1 × (16.24/1000).

4. We know that the volume of HCl used is 20ml, which is equivalent to 0.02 L. From this information, we can determine the concentration of HCl in g/L.
- Concentration = moles / volume in liters.
- Concentration = (0.1 × (16.24/1000)) / 0.02 moles per liter.
- Concentration = 0.0812 moles per liter.

C) Calculate the molarity of HCl:
1. The molarity of a solution is defined as the number of moles of solute per liter of solution.

2. Since the concentration we found in part B is already in moles per liter, it is the same as the molarity of HCl.
- Molarity of HCl = 0.0812 moles per liter.

Therefore, the answers are:
A) The amount of K2CO3 in the mixture is 1/425 moles per liter.
B) The concentration of HCl in g/L is 0.0812 moles per liter.
C) The molarity of HCl is 0.0812 moles per liter.

I hope this explanation was helpful! Let me know if you have any further questions.