1g of mixture na2c03 and k2c03 was mole made upto 250ml is aqueous solution. 25ml of this solution was neutralized. 20ml of hcl of unknown concentration. The neutralized solution required 16.24ml of 0.1N AGNo3 for precipitation, Calculate a) the % k2c03 is mixture b) conc of hcl in g/Litre c) molarity of hcl

Who helps me step by step

I try to tell you I am no chemist !

This sentence does not make sense to me:

1g of mixture na2c03 and k2c03 was mole made upto 250ml is aqueous solution.

What does "mole made up" mean?

Please also check the rest for typing errors.

Na2CO3 =2(23)+12+3(16) = 106 g/mol

x moles of Na2 CO3 mass = 106 x grams

K2CO3 = 2(39)+12+3(16) = 138 g/mol
y moles of K2CO3 mass = 138 y grams

106 x + 138 y = 1 gram total

in the titration I used 25 mL of my 250 ml so I used 1/10 gram solids of my original carbonates mixture

The reactions are
Na2CO3 + 2HCl --> 2NaCl + H2O + CO2
K2CO3 + 2HCl --> 2 KCl + H2O + CO2

Not being a chemist I will assume that I get the same number of moles of NaCl plus KCl as I had moles of AgNO3
How many is that?
moles AgNO3 = .1moles/liter * .01624 L = .001624 moles of NO3-
so I had .001624 moles of NaCl plus KCl
which means I had .000812 moles of
Na2CO3 plus K2CO3
x/10 moles Na2CO3 + y/10moles K2NO3 =.000812 moles
or
x + y = .00812
and from way long ago
106 x + 138 y = 1 gram total
---------------------
106 (.00812-y) + 138 y = 1

.861 -106 y + 138 y = 1
32 y = .139
y = .00435
fraction of y which is K2CO3 = .00435/.00812 = .536 which is 53.6 %

Wait a minute, x and y are moles and you want percentage of grams perhaps.

x = .00377
108 x = .40716
138 y = .6003

so yes, 60% K2CO3

Sure, I'd be happy to help you step by step!

a) To calculate the percentage of K2CO3 in the mixture, we need to first determine the moles of K2CO3 and Na2CO3 in the solution.

1) Let's calculate the number of moles of AgNO3 required for precipitation:
Given that 16.24 ml of 0.1N AgNO3 is required for the neutralization, we can calculate the number of moles of AgNO3 used:
Moles of AgNO3 = Volume (in liters) x Normality = 16.24 ml x 0.1 N / 1000 ml/L = 0.001624 moles

2) Now, let's calculate the number of moles of CO3^2- ions present in the solution:
Each mole of AgNO3 reacts with two moles of CO3^2- ions in the reaction:
Moles of CO3^2- ions = 0.001624 moles (since we used this amount of AgNO3)

3) Since Na2CO3 and K2CO3 are mixed in a 1:1 ratio, the number of moles of CO3^2- ions will be equal to the sum of the moles of Na2CO3 and K2CO3:
Moles of CO3^2- ions = Moles of Na2CO3 + Moles of K2CO3

4) Since we have a 1:1 ratio, the moles of Na2CO3 will be the same as the moles of CO3^2- ions:
Moles of Na2CO3 = Moles of CO3^2- ions = 0.001624 moles (from step 3)

5) Finally, let's calculate the percentage of K2CO3 in the mixture:
Since the mixture is made up of Na2CO3 and K2CO3, the total moles of the mixture will be equal to the sum of the moles of Na2CO3 and K2CO3:
Total moles of mixture = Moles of Na2CO3 + Moles of K2CO3 = 0.001624 moles (from step 4)

The percentage of K2CO3 in the mixture can be calculated using the following formula:
Percentage of K2CO3 = (Moles of K2CO3 / Total moles of mixture) x 100

b) To calculate the concentration of HCl in g/Litre:
We know that 20 ml of HCl of unknown concentration is used for neutralization.

1) First, we need to convert the volume of HCl used (20 ml) into liters:
Volume of HCl used = 20 ml = 20 ml / 1000 ml/L = 0.02 L

2) Next, we need to determine the number of moles of HCl used for neutralization. Since HCl is a strong acid, it fully dissociates in water, giving one mole of H+ ions for every mole of HCl:
Moles of HCl = Moles of H+ ions = 0.02 L (since the number of moles of H+ ions is equal to the volume of HCl used in liters)

3) Since the concentration is in g/Litre, we need to calculate the mass of HCl in grams:
Mass of HCl = Moles of HCl x Molar mass of HCl
The molar mass of HCl is approximately 36.5 g/mol.

c) To calculate the molarity of HCl:
Molarity (M) = Moles of HCl / Volume in liters

Please provide me with the molar masses of Na2CO3 and K2CO3, and the molar mass of HCl if you have them, or I can assume approximate values.