1g of mixture of na2c03 and k2c03 was made upto 250ml is aqueous solution . 25ml of this solution was neutralized 20ml of hcl of unknow concentration. The neutralized solution required 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
A is answer 60%
B is answer 2.9g/L
c) is answer 0.0812M
Damon wants you step by step for me.
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 can guide you through the steps to solve this problem. Let's break it down step by step:
Given information:
- A mixture of Na2CO3 and K2CO3 with a total mass of 1g was made into a solution of 250ml.
- 25ml of this solution required 20ml of HCl of unknown concentration for neutralization.
- The neutralized solution required 16.24ml of 0.1N AgNO3 for precipitation.
a) To calculate the percentage of K2CO3 in the mixture:
1. Determine the moles of AgNO3 used for precipitation.
Number of moles of AgNO3 = Molarity of AgNO3 × Volume of AgNO3 used (in liters)
Number of moles of AgNO3 = 0.1N × 0.01624L (convert ml to L)
Note: 0.1N = 0.1 moles/L
2. Since AgNO3 precipitates with K2CO3 but not with Na2CO3, the moles of AgNO3 used will be equal to the moles of K2CO3 in the mixture.
Number of moles of K2CO3 = Number of moles of AgNO3
3. Determine the molar mass of K2CO3:
Molar mass of K2CO3 = (2 × atomic mass of K) + atomic mass of C + (3 × atomic mass of O)
4. Calculate the mass of K2CO3:
Mass of K2CO3 = Number of moles of K2CO3 × Molar mass of K2CO3
5. Calculate the percentage of K2CO3 in the mixture:
Percentage of K2CO3 = (Mass of K2CO3 / Total mass of the mixture) × 100
b) To calculate the concentration of HCl in g/L:
1. Determine the number of moles of HCl used for neutralization.
Number of moles of HCl = Molarity of HCl × Volume of HCl used (in liters)
Volume of HCl used = 20ml = 0.02L (convert ml to L)
Note: The concentration of HCl in g/L is the same as the molarity.
2. Calculate the concentration of HCl in g/L:
Concentration of HCl (g/L) = (Number of moles of HCl / Volume of solution) × 1000
c) To calculate the molarity of HCl:
1. Determine the molarity of AgNO3 used for precipitation.
Molarity of AgNO3 = 0.1N
2. Since AgNO3 reacts with Cl- ions in HCl in a 1:1 ratio, the number of moles of AgNO3 used will be equal to the number of moles of Cl- ions in the solution.
Number of moles of Cl- = Number of moles of AgNO3
3. The volume of the neutralized solution used for precipitation is 25ml, which is the same as 0.025L (convert ml to L).
4. Calculate the molarity of HCl:
Molarity of HCl = (Number of moles of Cl- / Volume of the solution in liters)
Following these steps, you should be able to calculate the answers for parts a), b), and c) of the question.